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Linear Algebra : Quadratic Forms and Positive Semidefinite Matrices

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Example Question #11 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{Use eigenvalues to determine if the following matrix is positive semi-definite:}\begin{bmatrix}17&16&-5\\16&17&-4\\-5&-4&16\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The matrix is not positive semi-definite.}$

$\text{The matrix is positive semi-definite.}$

Correct answer:

$\text{The matrix is positive semi-definite.}$

Explanation:

$\begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}17&16&-5\\16&17&-4\\-5&-4&16\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}17 - \lambda&16&-5\\16&17 - \lambda&-4\\-5&-4&16 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&(-1)\cdot (536\lambda - 50\lambda ^{2} + \lambda ^{3} - 471)=0\\&\lambda_{1}=0.96;\lambda_{2}=13.92;\lambda_{3}=35.12\\&\text{The matrix is positive semi-definite.}\end{align*}$

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Example Question #12 : Quadratic Forms And Positive Semidefinite Matrices

$\begin{align*}&\text{Decide whether or not the matrix }\begin{bmatrix}15&-11&8\\-11&20&-1\\8&-1&10\end{bmatrix}\\&\text{is positive semi-definite using eigenvalues.}\end{align*}$

Possible Answers:

$\text{The matrix is not positive semi-definite.}$

$\text{The matrix is positive semi-definite.}$

Correct answer:

$\text{The matrix is positive semi-definite.}$

Explanation:

$\begin{align*}&\text{The matrix given in the problem is symmetric; that is to say,}\\&\text{it is identical to its transpose. If a matrix is symmetric,}\\&\text{then it is also positive semi-definite if all of its eigenvalues}\\&\text{are non-negative. Computing these eigenvalues:}\\&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)=0\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}15&-11&8\\-11&20&-1\\8&-1&10\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}15 - \lambda&-11&8\\-11&20 - \lambda&-1\\8&-1&10 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&(-1)\cdot (464\lambda - 45\lambda ^{2} + \lambda ^{3} - 671)=0\\&\lambda_{1}=1.72;\lambda_{2}=12.76;\lambda_{3}=30.51\\&\text{The matrix is positive semi-definite.}\end{align*}$

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Example Question #11 : Quadratic Forms And Positive Semidefinite Matrices

$D = \begin{bmatrix} 4 & \log_{3} 4 & \log_{3}4 & \log_{3}4 \\ \log_{5}6 & 4 & \log_{5}6& \log_{5} 6 \\ \log_{7} 8 & \log_{7}8 & 4 & \log_{7}8 \\ \log_{9} 10 & \log_{9}10 & \log_{9}10 & 4 \end{bmatrix}$

True or false: is an example of a diagonally dominant matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

A real matrix is a diagonally dominant matrix if and only if, in each row, the absolute value of the diagonal element is greater than or equal to the sum of the absolute values of the other elements in that row. Below is with the diagonal elements in red:

$D = \begin{bmatrix}{\color{Red} \textbf{4} }& \log_{3} 4 & \log_{3}4 & \log_{3}4 \\ \log_{5}6 & {\color{Red} \textbf{4} } & \log_{5}6& \log_{5} 6 \\ \log_{7} 8 & \log_{7}8 & {\color{Red} \textbf{4} } & \log_{7}8 \\ \log_{9} 10 & \log_{9}10 & \log_{9}10 & {\color{Red} \textbf{4} } \end{bmatrix}$

For each row, compare the absolute value of the diagonal element to the sum of the absolute values of the other elements. Note that each nondiagonal element is a logarithm, with base greater than 1, of a number greater than 1, so each element is positive; also, to calculate each nondiagonal sum, the sum rule for logarithms will be employed:

Row 1:

$\log_{3}4 + \log_{3}4 + \log_{3}4 = \log_{3} (4 \cdot 4 \cdot 4 )= \log_{3} 64$

$4 = \log_{3}(3^{4})= \log_{3} 81$

Therefore,

$|4 |>|\log_{3}4| +|\log_{3}4| +|\log_{3}4|$ .

Row 2:

$\log_{5}6 +\log_{5}6 +\log_{5}6 = \log_{5} (6 \cdot 6 \cdot 6 )= \log_{5} 216$

$4 = \log_{5}(5^{4}) =\log_{5}625$

Therefore,

$|4 |>|\log_{5}6| +|\log_{5}6| +|\log_{5}6|$ .

Row 3:

$\log_{7} 8 + \log_{7} 8+ \log_{7} 8 = \log_{7}( 8 \cdot 8 \cdot 8 )= \log_{7} 512$

$4 = \log_{7}(7^{4})= \log_{5}2,401$

Therefore,

$|4 |>|\log_{7} 8 | + |\log_{7} 8 | + |\log_{7} 8 |$ .

Row 4:

$\log_{9}10+ \log_{9}10 + \log_{9}10 = \log_{9}( 10 \cdot 10 \cdot 10 )= \log_{9}1,000$

$4 = \log_{9}(9^{4}) =\log_{9} 6,561$

Therefore,

$|4 |>| \log_{9}10| + | \log_{9}10| + | \log_{9}10 |$ .

In each row, the absolute value of the diagonal element is greater than or equal to the sum of the absolute values of the other elements. is diagonally dominant.

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Example Question #14 : Quadratic Forms And Positive Semidefinite Matrices

$D = \begin{bmatrix} -e^{2} & e & 1 \\ e & -e^{3} & e^{2} \\ e^{3}& e & -e^{4} \end{bmatrix}$

True or false: is an example of a diagonally dominant matrix.

Possible Answers:

True

False

Correct answer:

True

Explanation:

$D = \begin{bmatrix} {\color{Red} \textbf{-7.3891}}& 2.7183 & 1 \\ 2.7183 & {\color{Red} \textbf{-20.0855}}& 7.3891 \\ 20.0855&2.7183 & {\color{Red} \textbf{-54.5982}} \end{bmatrix}$

For each row, compare the absolute value of the diagonal element to the sum of the absolute values of the other elements.

Row 1:

|2.7183| + |1| = 2.7183 + 1 = 3.7183 < 7.3891 = |-7.3891|

Row 2:

| 2.7183| + | 7.3891| = 2.7183+ 7.3891 = 10.1074 <20.0855 =|-20.0855|

Row 3:

|20.0855|+ |2.7183 |= 20.0855+ 2.7183 =22.8038 < 54.5982 =| -54.5982 |

In each row, the absolute value of the diagonal element is greater than or equal to the sum of the absolute values of the other elements. is diagonally dominant.

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Example Question #11 : Quadratic Forms And Positive Semidefinite Matrices

Consider the diagonally dominant matrix

$A = \begin{bmatrix} 20 & 0.1 & 0.2 & -0.3 & 0 \\ 0.2 & -15 & 0.4 & 0.1 & -0.3 \\ 0.3 & -0.2 & 18 & -0.2 & 0.2 \\ 0.2 & - 0.2 & 0.5 & 16 & 0.2 \\ 0.4 & 0 & 0.1 & 0.8 & -40 \end{bmatrix}$

Of the five Gershgorin discs of , give the center of the one with the greatest radius.

Possible Answers:

Correct answer:

Explanation:

The Gershgorin discs of a diagonally dominant matrix have as their centers the diagonal elements of the matrix. The radius of a disc is the sum of the absolute values of the nondiagonal elements in the same row as the given diagonal element. Therefore, note the diagonal elements of :

$A = \begin{bmatrix} \textbf{20 }& 0.1 & 0.2 & -0.3 & 0 \\ 0.2 & \textbf{-15}& 0.4 & 0.1 & -0.3 \\ 0.3 & -0.2 & \textbf{18}& -0.2 & 0.2 \\ 0.2 & - 0.2 & 0.5 & \textbf{16}& 0.2 \\ 0.4 & 0 & 0.1 & 0.8 & \textbf{-40} \end{bmatrix}$

The sums of the absolute values of the nondiagonal elements corresponding to each diagonal element are:

20:|0.1|+ |0.2|+|-0.3| + |0|= 0.6

-15:| 0.2| + |0.4| + | 0.1| + | -0.3| = 1.0

18: |0.3|+ | -0.2| +| -0.2| +| 0.2| = 0.9

16: | 0.2| + |- 0.2| + | 0.5| +| 0.2| = 1.1

$-40: | 0.4|+ |0| +| 0.1| +|0.8| = \textbf{{\color{Red}1.3}}$

The largest of the Gershgorin discs is the one with radius 1.3, with center .

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Example Question #11 : Quadratic Forms And Positive Semidefinite Matrices

Consider the diagonally dominant matrix

$A = \begin{bmatrix} 20 & 0.1 & 0.2 & -0.3 & 0 \\ 0.2 & -15 & 0.4 & 0.1 & -0.3 \\ 0.3 & -0.2 & 18 & -0.2 & 0.2 \\ 0.2 & - 0.2 & 0.5 & 16 & 0.2 \\ 0.4 & 0 & 0.1 & 0.8 & -40 \end{bmatrix}$

Of the five Gershgorin discs of , give the center of the one with the least radius.

Possible Answers:

Correct answer:

Explanation:

The sums of the absolute values of the nondiagonal elements corresponding to each diagonal element are:

$20:|0.1|+ |0.2|+|-0.3| + |0|= \textbf{{\color{Red} 0.6}}$

-15:| 0.2| + |0.4| + | 0.1| + | -0.3| = 1.0

18: |0.3|+ | -0.2| +| -0.2| +| 0.2| = 0.9

16: | 0.2| + |- 0.2| + | 0.5| +| 0.2| = 1.1

-40: | 0.4|+ |0| +| 0.1| +|0.8| = 1.3

The smallest of the Gershgorin discs is the one with radius 0.6, with center 20.

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Example Question #17 : Quadratic Forms And Positive Semidefinite Matrices

$A = \begin{bmatrix} 20 & 0.1 & 0.2 & -0.3 & 0 \\ 0.2 & -15 & 0.4 & 0.1 & -0.3 \\ 0.3 & -0.2 & 18 & -0.2 & 0.2 \\ 0.2 & - 0.2 & 0.5 & 16 & 0.2 \\ 0.4 & 0 & 0.1 & 0.8 & -40 \end{bmatrix}$

True or false: is an example of a diagonally dominant matrix.

Possible Answers:

True

False

Correct answer:

True

Explanation:

A matrix is diagonally dominant if, in each row, the absolute value of the diagonal element is greater than or equal to the sum of the absolute values of the other elements. Compare the quantities in each row:

$A = \begin{bmatrix} 20 & 0.1 & 0.2 & -0.3 & 0 \\ 0.2 & -15 & 0.4 & 0.1 & -0.3 \\ 0.3 & -0.2 & 18 & -0.2 & 0.2 \\ 0.2 & - 0.2 & 0.5 & 16 & 0.2 \\ 0.4 & 0 & 0.1 & 0.8 & -40 \end{bmatrix}$

20 = |20| > |0.1|+ |0.2|+|-0.3| + |0|= 0.6

15 = |-15 |> | 0.2| + |0.4| + | 0.1| + | -0.3| = 1.0

18 = |18| > |0.3|+ | -0.2| +| -0.2| +| 0.2| = 0.9

16 = |16|> | 0.2| + |- 0.2| + | 0.5| +| 0.2| = 1.1

40 = |- 40|> | 0.4|+ |0| +| 0.1| +|0.8| = 1.3

In each case, the condition holds, so is diagonally dominant.

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Example Question #18 : Quadratic Forms And Positive Semidefinite Matrices

Consider the diagonally dominant matrix

$A = \begin{bmatrix} 6 & \frac{3}{5} \\ \frac{2}{3} & 7 \end{bmatrix}$

True or false: The Gershgorin discs of intersect.

Possible Answers:

False

True

Correct answer:

True

Explanation:

On the complex plane, the Gershgorin discs of a diagonally dominant matrix are the circles with centers at the diagonal elements, whose radii are equal to the sums of the nondiagonal elements in the same row. The two Gershgorin discs of are the circle with center 6 and radius $\frac{3}{5}$ , and the circle with center 7 and radius $\frac{2}{3}$ , as seen below:

$A = \begin{bmatrix} {\color{Blue} \textbf{6} }&{\color{Red} \frac{ \textbf{3}}{ \textbf{5}}} \\ {\color{Red} \frac{ \textbf{2}}{ \textbf{3}}} & {\color{Blue} \textbf{7}} \end{bmatrix}$

The centers of the discs are 1 unit apart. The total of the radii of the discs is

$\frac{3}{5}+ \frac{2}{3}= \frac{19}{15} > 1$ , so the sum of the radii is greater than the distance between the centers. The discs intersect.

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Linear Algebra : Quadratic Forms and Positive Semidefinite Matrices

Study concepts, example questions & explanations for Linear Algebra

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Example Question #11 : Quadratic Forms And Positive Semidefinite Matrices

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Example Question #11 : Quadratic Forms And Positive Semidefinite Matrices

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