Special types of matrices

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Special types of matrices

Definition 10

A matrixA ∈ R^n×n is said to bestrictly diagonally dominantif

|a_ii| >

n

X

j=1,j6=i

|a_ij|.

Lemma 11

IfA ∈ R^n×n isstrictly diagonally dominant, thenAis nonsingular.

Proof: Suppose A is singular. Then there exists x ∈ Rⁿ, x 6= 0 such that Ax = 0.Let k be the integer index such that

|x_k| = max

1≤i≤n|x_i| =⇒ |x_i|

|x_k| ≤ 1, ∀ |x_i|.

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Special types of matrices

Definition 10

A matrixA ∈ R^n×n is said to bestrictly diagonally dominantif

|a_ii| >

n

X

j=1,j6=i

|a_ij|.

Lemma 11

IfA ∈ R^n×n isstrictly diagonally dominant, thenAis nonsingular.

Proof: Suppose A is singular.Then there exists x ∈ Rⁿ, x 6= 0 such that Ax = 0. Let k be the integer index such that

|x_k| = max

1≤i≤n|x_i| =⇒ |x_i|

|x_k| ≤ 1, ∀ |x_i|.

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Special types of matrices

Definition 10

A matrixA ∈ R^n×n is said to bestrictly diagonally dominantif

|a_ii| >

n

X

j=1,j6=i

|a_ij|.

Lemma 11

IfA ∈ R^n×n isstrictly diagonally dominant, thenAis nonsingular.

Proof: Suppose A is singular. Then there exists x ∈ Rⁿ, x 6= 0 such that Ax = 0.Let k be the integer index such that

|x_k| = max

1≤i≤n|x_i| =⇒ |x_i|

|x_k| ≤ 1, ∀ |x_i|.

163 / 255

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Special types of matrices

Definition 10

A matrixA ∈ R^n×n is said to bestrictly diagonally dominantif

|a_ii| >

n

X

j=1,j6=i

|a_ij|.

Lemma 11

IfA ∈ R^n×n isstrictly diagonally dominant, thenAis nonsingular.

Proof: Suppose A is singular. Then there exists x ∈ Rⁿ, x 6= 0 such that Ax = 0. Let k be the integer index such that

|x_k| = max

1≤i≤n|x_i| =⇒ |x_i|

|x_k| ≤ 1, ∀ |x_i|.

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Since Ax = 0, for the fixed k, we have

n

X

j=1

a_kjxj = 0 ⇒ a_kkx_k= −

n

X

j=1,j6=k

a_kjxj

⇒ |a_kk||x_k| ≤

n

X

j=1,j6=k

|a_kj||x_j|,

which implies

|a_kk| ≤

n

X

j=1,j6=k

|a_kj||x_j|

|x_k|≤

n

X

j=1,j6=k

|a_kj|.

But this contradicts the assumption that A is diagonally dominant.Therefore A must be nonsingular.

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Since Ax = 0, for the fixed k, we have

n

X

j=1

a_kjxj = 0 ⇒ a_kkx_k= −

n

X

j=1,j6=k

a_kjxj

⇒ |a_kk||x_k| ≤

n

X

j=1,j6=k

|a_kj||x_j|,

which implies

|a_kk| ≤

n

X

j=1,j6=k

|a_kj||x_j|

|x_k|≤

n

X

j=1,j6=k

|a_kj|.

But this contradicts the assumption that A is diagonally dominant. Therefore A must be nonsingular.

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Since Ax = 0, for the fixed k, we have

n

X

j=1

a_kjxj = 0 ⇒ a_kkx_k= −

n

X

j=1,j6=k

a_kjxj

⇒ |a_kk||x_k| ≤

n

X

j=1,j6=k

|a_kj||x_j|,

which implies

|a_kk| ≤

n

X

j=1,j6=k

|a_kj||x_j|

|x_k|≤

n

X

j=1,j6=k

|a_kj|.

But this contradicts the assumption that A is diagonally dominant.Therefore A must be nonsingular.

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Since Ax = 0, for the fixed k, we have

n

X

j=1

a_kjxj = 0 ⇒ a_kkx_k= −

n

X

j=1,j6=k

a_kjxj

⇒ |a_kk||x_k| ≤

n

X

j=1,j6=k

|a_kj||x_j|,

which implies

|a_kk| ≤

n

X

j=1,j6=k

|a_kj||x_j|

|x_k|≤

n

X

j=1,j6=k

|a_kj|.

But this contradicts the assumption that A is diagonally dominant. Therefore A must be nonsingular.

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Theorem 12

Gaussian elimination without pivoting preserve the diagonal dominance of a matrix.

Proof: Let A ∈ R^n×n be a diagonally dominant matrix and A⁽²⁾= [a⁽²⁾_ij ]is the result of applying one step of Gaussian elimination to A⁽¹⁾= Awithout any pivoting strategy.

After one step of Gaussian elimination, a⁽²⁾_i1 = 0for i = 2, . . . , n, and the first row is unchanged.Therefore, the property

|a⁽²⁾₁₁| >

n

X

j=2

|a⁽²⁾_1j |

is preserved, and all we need to show is that

|a⁽²⁾_ii | >

n

X

j=2,j6=i

|a⁽²⁾_ij |, for i = 2, . . . , n.

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Theorem 12

Gaussian elimination without pivoting preserve the diagonal dominance of a matrix.

Proof: Let A ∈ R^n×n be a diagonally dominant matrix and A⁽²⁾= [a⁽²⁾_ij ]is the result of applying one step of Gaussian elimination to A⁽¹⁾= Awithout any pivoting strategy.

After one step of Gaussian elimination, a⁽²⁾_i1 = 0for i = 2, . . . , n, and the first row is unchanged. Therefore, the property

|a⁽²⁾₁₁| >

n

X

j=2

|a⁽²⁾_1j |

is preserved,and all we need to show is that

|a⁽²⁾_ii | >

n

X

j=2,j6=i

|a⁽²⁾_ij |, for i = 2, . . . , n.

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Theorem 12

Gaussian elimination without pivoting preserve the diagonal dominance of a matrix.

Proof: Let A ∈ R^n×n be a diagonally dominant matrix and A⁽²⁾= [a⁽²⁾_ij ]is the result of applying one step of Gaussian elimination to A⁽¹⁾= Awithout any pivoting strategy.

After one step of Gaussian elimination, a⁽²⁾_i1 = 0for i = 2, . . . , n, and the first row is unchanged.Therefore, the property

|a⁽²⁾₁₁| >

n

X

j=2

|a⁽²⁾_1j |

is preserved, and all we need to show is that

|a⁽²⁾_ii | >

n

X

j=2,j6=i

|a⁽²⁾_ij |, for i = 2, . . . , n.

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Theorem 12

Gaussian elimination without pivoting preserve the diagonal dominance of a matrix.

Proof: Let A ∈ R^n×n be a diagonally dominant matrix and A⁽²⁾= [a⁽²⁾_ij ]is the result of applying one step of Gaussian elimination to A⁽¹⁾= Awithout any pivoting strategy.

After one step of Gaussian elimination, a⁽²⁾_i1 = 0for i = 2, . . . , n, and the first row is unchanged. Therefore, the property

|a⁽²⁾₁₁| >

n

X

j=2

|a⁽²⁾_1j |

is preserved,and all we need to show is that

|a⁽²⁾_ii | >

n

X

j=2,j6=i

|a⁽²⁾_ij |, for i = 2, . . . , n.

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Theorem 12

Gaussian elimination without pivoting preserve the diagonal dominance of a matrix.

Proof: Let A ∈ R^n×n be a diagonally dominant matrix and A⁽²⁾= [a⁽²⁾_ij ]is the result of applying one step of Gaussian elimination to A⁽¹⁾= Awithout any pivoting strategy.

After one step of Gaussian elimination, a⁽²⁾_i1 = 0for i = 2, . . . , n, and the first row is unchanged. Therefore, the property

|a⁽²⁾₁₁| >

n

X

j=2

|a⁽²⁾_1j |

is preserved, and all we need to show is that

|a⁽²⁾_ii | >

n

X

j=2,j6=i

|a⁽²⁾_ij |, for i = 2, . . . , n.

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Using the Gaussian elimination formula (2), we have

|a⁽²⁾_ii | =     

a⁽¹⁾_ii −a⁽¹⁾_i1 a⁽¹⁾₁₁

a⁽¹⁾_1i     

=    

aii−ai1

a₁₁a1i

   

≥ |aii| − |ai1|

|a11||a1i|

= |aii| − |ai1| + |ai1| − |a_i1|

|a11||a1i|

= |aii| − |ai1| + |ai1|

|a11|(|a11| − |a1i|)

>

n

X

j=2,j6=i

|a_ij| + |a_i1|

|a11|

n

X

j=2,j6=i

|a_1j|

=

n

X

j=2,j6=i

|aij| +

n

X

j=2,j6=i

|ai1|

|a11||a1j|

≥

n

X

j=2,j6=i

   

aij− ai1

a11

a1j

   

=

n

X

j=2,j6=i

|a⁽²⁾_ij |.

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Thus A⁽²⁾is still diagonally dominant.Since the subsequent steps of Gaussian elimination mimic the first, except for being applied to submatrices of smaller size, it suffices to conclude that Gaussian elimination without pivoting preserves the diagonal dominance of a matrix.

Theorem 13

Let A be strictly diagonally dominant. Then Gaussian elimination can be performed on Ax = b to obtain its unique solution without row or column interchanges.

Definition 14

A matrix A ispositive definiteif it issymmetricandx^TAx > 0

∀ x 6= 0.

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Thus A⁽²⁾is still diagonally dominant. Since the subsequent steps of Gaussian elimination mimic the first, except for being applied to submatrices of smaller size, it suffices to conclude that Gaussian elimination without pivoting preserves the diagonal dominance of a matrix.

Theorem 13

Let A be strictly diagonally dominant. Then Gaussian elimination can be performed on Ax = b to obtain its unique solution without row or column interchanges.

Definition 14

A matrix A ispositive definiteif it issymmetricandx^TAx > 0

∀ x 6= 0.

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Thus A⁽²⁾is still diagonally dominant. Since the subsequent steps of Gaussian elimination mimic the first, except for being applied to submatrices of smaller size, it suffices to conclude that Gaussian elimination without pivoting preserves the diagonal dominance of a matrix.

Theorem 13

Let A be strictly diagonally dominant. Then Gaussian elimination can be performed on Ax = b to obtain its unique solution without row or column interchanges.

Definition 14

A matrix A ispositive definiteif it issymmetricandx^TAx > 0

∀ x 6= 0.

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Thus A⁽²⁾is still diagonally dominant. Since the subsequent steps of Gaussian elimination mimic the first, except for being applied to submatrices of smaller size, it suffices to conclude that Gaussian elimination without pivoting preserves the diagonal dominance of a matrix.

Theorem 13

Let A be strictly diagonally dominant. Then Gaussian elimination can be performed on Ax = b to obtain its unique solution without row or column interchanges.

Definition 14

A matrix A ispositive definiteif it issymmetricandx^TAx > 0

∀ x 6= 0.

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0. Since A is positive definite, this implies x = 0. Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 179 / 255

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0. Since A is positive definite, this implies x = 0. Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 180 / 255

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0. Since A is positive definite, this implies x = 0. Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 181 / 255

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0.Since A is positive definite, this implies x = 0. Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 182 / 255

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0. Since A is positive definite, this implies x = 0.Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 183 / 255

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0.Since A is positive definite, this implies x = 0. Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 184 / 255

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0. Since A is positive definite, this implies x = 0.Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 185 / 255

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0. Since A is positive definite, this implies x = 0. Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 186 / 255

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Theorem 15

If A is an n × n positive definite matrix, then (a) Ahas an inverse;

(b) a_ii> 0, ∀ i = 1, . . . , n;

(c) max_1≤k,j≤n|a_kj| ≤ max_1≤i≤n|a_ii|;

(d) (a_ij)² < a_iia_jj, ∀ i 6= j.

Proof:

(a) If x satisfies Ax = 0, then x^TAx = 0. Since A is positive definite, this implies x = 0. Consequently, Ax = 0has only the zero solution, and A is nonsingular.

(b) Since A is positive definite, a_ii= e^T_i Ae_i> 0,

where ei is the i-th column of the n × n identify

matrix. 187 / 255

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(c) For k 6= j, define x = [x_i]by

xi =







0, if i 6= j and i 6= k, 1, if i = j,

−1, if i = k.

Since x 6= 0,

0 < x^TAx = a_jj+ a_kk− a_jk− a_kj. But A^T = A, so

2akj < ajj+ akk. (5) Now define z = [z_i]by

z_i =

 0, if i 6= j and j 6= k, 1, if i = j or i = k.

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(c) For k 6= j, define x = [x_i]by

xi =







0, if i 6= j and i 6= k, 1, if i = j,

−1, if i = k.

Since x 6= 0,

0 < x^TAx = a_jj+ a_kk− a_jk− a_kj. But A^T = A, so

2akj < ajj+ akk. (5) Now define z = [z_i]by

z_i =

 0, if i 6= j and j 6= k, 1, if i = j or i = k.

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(c) For k 6= j, define x = [x_i]by

xi =







0, if i 6= j and i 6= k, 1, if i = j,

−1, if i = k.

Since x 6= 0,

0 < x^TAx = a_jj+ a_kk− a_jk− a_kj. But A^T = A, so

2akj < ajj+ akk. (5) Now define z = [z_i]by

z_i =

 0, if i 6= j and j 6= k, 1, if i = j or i = k.

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(c) For k 6= j, define x = [x_i]by

xi =







0, if i 6= j and i 6= k, 1, if i = j,

−1, if i = k.

Since x 6= 0,

0 < x^TAx = a_jj+ a_kk− a_jk− a_kj. But A^T = A, so

2akj < ajj+ akk. (5) Now define z = [z_i]by

z_i =

 0, if i 6= j and j 6= k, 1, if i = j or i = k.

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Then z^TAz > 0, so

−2a_kj < a_jj+ a_kk. (6) Equations (5) and (6) imply that for each k 6= j,

|a_kj| < a_kk+ a_jj

2 ≤ max

1≤i≤n|a_ii|, so

1≤k,j≤nmax |a_kj| ≤ max

1≤i≤n|a_ii|.

(d) For i 6= j, define x = [x_k]by

xk=







0, if k 6= j and k 6= i, α, if k = i,

1, if k = j,

where α represents an arbitrary real number.

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Then z^TAz > 0, so

−2a_kj < a_jj+ a_kk. (6) Equations (5) and (6) imply that for each k 6= j,

|a_kj| < a_kk+ a_jj

2 ≤ max

1≤i≤n|a_ii|, so

1≤k,j≤nmax |a_kj| ≤ max

1≤i≤n|a_ii|.

(d) For i 6= j, define x = [x_k]by

xk=







0, if k 6= j and k 6= i, α, if k = i,

1, if k = j,

where α represents an arbitrary real number.

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Then z^TAz > 0, so

−2a_kj < a_jj+ a_kk. (6) Equations (5) and (6) imply that for each k 6= j,

|a_kj| < a_kk+ a_jj

2 ≤ max

1≤i≤n|a_ii|, so

1≤k,j≤nmax |a_kj| ≤ max

1≤i≤n|a_ii|.

(d) For i 6= j, define x = [x_k]by

xk=







0, if k 6= j and k 6= i, α, if k = i,

1, if k = j,

where α represents an arbitrary real number.

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Then z^TAz > 0, so

−2a_kj < a_jj+ a_kk. (6) Equations (5) and (6) imply that for each k 6= j,

|a_kj| < a_kk+ a_jj

2 ≤ max

1≤i≤n|a_ii|, so

1≤k,j≤nmax |a_kj| ≤ max

1≤i≤n|a_ii|.

(d) For i 6= j, define x = [x_k]by

xk=







0, if k 6= j and k 6= i, α, if k = i,

1, if k = j,

where α represents an arbitrary real number.

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Since x 6= 0,

0 < x^TAx = aiiα²+ 2aijα + ajj ≡ P (α), ∀ α ∈ R.

That is the quadratic polynomial P (α) has no real roots. It implies that

4a²_ij − 4a_iiajj < 0 and a²_ij < aiiajj. Definition 16 (Leading principal minor)

Let A be an n × n matrix. Theupper leftk × ksubmatrix, denoted as

A_k=











a₁₁ a₁₂ · · · a_1k a21 a22 · · · a2k

... ... . .. ... ak1 ak2 · · · akk









 ,

is called theleading k × k principal submatrix, and the determinant of A_k,det(A_k), is called theleading principal

minor. ^{196 / 255}

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Since x 6= 0,

0 < x^TAx = aiiα²+ 2aijα + ajj ≡ P (α), ∀ α ∈ R.

That is the quadratic polynomial P (α) has no real roots.It implies that

4a²_ij − 4a_iiajj < 0 and a²_ij < aiiajj. Definition 16 (Leading principal minor)

Let A be an n × n matrix. Theupper leftk × ksubmatrix, denoted as

A_k=











a₁₁ a₁₂ · · · a_1k a21 a22 · · · a2k

... ... . .. ... ak1 ak2 · · · akk









 ,

is called theleading k × k principal submatrix, and the determinant of A_k,det(A_k), is called theleading principal

minor. ^{197 / 255}

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Since x 6= 0,

0 < x^TAx = aiiα²+ 2aijα + ajj ≡ P (α), ∀ α ∈ R.

That is the quadratic polynomial P (α) has no real roots. It implies that

4a²_ij − 4a_iiajj < 0 and a²_ij < aiiajj. Definition 16 (Leading principal minor)

Let A be an n × n matrix. Theupper leftk × ksubmatrix, denoted as

A_k=











a₁₁ a₁₂ · · · a_1k a21 a22 · · · a2k

... ... . .. ... ak1 ak2 · · · akk









 ,

is called theleading k × k principal submatrix, and the determinant of A_k,det(A_k), is called theleading principal

minor. ^{198 / 255}

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Since x 6= 0,

0 < x^TAx = aiiα²+ 2aijα + ajj ≡ P (α), ∀ α ∈ R.

That is the quadratic polynomial P (α) has no real roots. It implies that

4a²_ij − 4a_iiajj < 0 and a²_ij < aiiajj. Definition 16 (Leading principal minor)

Let A be an n × n matrix. Theupper leftk × ksubmatrix, denoted as

A_k=











a₁₁ a₁₂ · · · a_1k a21 a22 · · · a2k

... ... . .. ... ak1 ak2 · · · akk









 ,

is called theleading k × k principal submatrix, and the determinant of A_k,det(A_k), is called theleading principal

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在文檔中 Direct Methods for Solving Linear Systems (頁 160-200)