Linear Algebra and Its Applications, Exercise 1.6.21

Exercise 1.6.21. Factor the following symmetric matrices

$A = \begin{bmatrix} 1&3&5 \\ 3&12&18 \\ 5&18&30 \end{bmatrix} \qquad A = \begin{bmatrix} a&b \\ b&d \end{bmatrix}$

into the form $LDL^T$.

Answer: We perform Gaussian elimination on the first matrix $A$ and start by multiplying the first row by the multiplier $l_{21} = 3$ and subtracting it from the second row:

$\begin{bmatrix} 1&3&5 \\ 3&12&18 \\ 5&18&30 \end{bmatrix} \rightarrow \begin{bmatrix} 1&3&5 \\ 0&3&3 \\ 5&18&30 \end{bmatrix}$

We then multiply the first row by the multiplier $l_{31} = 5$ and subtract it from the third row:

$\begin{bmatrix} 1&3&5 \\ 0&3&3 \\ 5&18&30 \end{bmatrix} \rightarrow \begin{bmatrix} 1&3&5 \\ 0&3&3 \\ 0&3&5 \end{bmatrix}$

Finally we multiply the second row by the multiplier $l_{32} = 1$ and subtract it from the third row:

$\begin{bmatrix} 1&3&5 \\ 0&3&3 \\ 0&3&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&3&5 \\ 0&3&3 \\ 0&0&2 \end{bmatrix}$

We then have

$L = \begin{bmatrix} 1&0&0 \\ l_{21}&1&0 \\ l_{31}&l_{32}&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 3&1&0 \\ 5&1&1 \end{bmatrix}$

(substituting for the $l_{ij}$ entries) and

$D = \begin{bmatrix} 1&0&0 \\ 0&3&0 \\ 0&0&2 \end{bmatrix}$

(taking the diagonal values from the matrix after the final elimination step).

The factorization of $A$ is then

$A = LDL^T = \begin{bmatrix} 1&0&0 \\ 3&1&0 \\ 5&1&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&3&0 \\ 0&0&2 \end{bmatrix} \begin{bmatrix} 1&3&5 \\ 0&1&1 \\ 0&0&1 \end{bmatrix}$

For the second matrix we also do Gaussian elimination by multiplying the first row by $l_{21} = b/a$ and subtracting it from the second:

$\begin{bmatrix} a&b \\ b&d \end{bmatrix} \rightarrow \begin{bmatrix} a&b \\ 0&d-b^2/a \end{bmatrix}$

This completes elimination. We then have

$L = \begin{bmatrix} 1&0 \\ l_{21}&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ b/a&1 \end{bmatrix}$

(substituting for the $l_{ij}$ entries) and

$D = \begin{bmatrix} a&0 \\ 0&d-b^2/a \end{bmatrix}$

(taking the diagonal values from the matrix after the final elimination step).

The factorization of $A$ is then

$A = LDL^T = \begin{bmatrix} 1&0 \\ b/a&1 \end{bmatrix} \begin{bmatrix} a&0 \\ 0&d-b^2/a \end{bmatrix} \begin{bmatrix} 1&b/a \\ 0&1 \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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