## 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 .

This entry was posted in linear algebra. Bookmark the permalink.