## Linear Algebra and Its Applications, Exercise 1.5.9

Exercise 1.5.9. (a) Assume that A is the product of three matrices as follows:

$A = \begin{bmatrix} 1&0&0 \\ -1&1&0 \\ 0&-1&1 \end{bmatrix} \begin{bmatrix} d_1&& \\ &d_2& \\ &&d_3 \end{bmatrix} \begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 0&0&1 \end{bmatrix}$

What must be true for A to be nonsingular?

(b) Given A above, assume we have a system of linear equations Ax = b with corresponding Lc = b as follows:

$\begin{bmatrix} 1&0&0 \\ -1&1&0 \\ 0&-1&1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = b$

Answer: (a) From the above we have A = LDU where L is a lower triangular matrix, D is a diagonal matrix, and U is an upper triangular matrix. Per p. 36 the diagonal entries of D are the pivots, and in order for A to be nonsingular all pivots must be nonzero. So A is nonsingular if and only if $d_1$, $d_2$, and $d_3$ are not zero.

(b) From above we have

$b = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ -1&1&0 \\ 0&-1&1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2-c_1 \\ c_3-c_2 \end{bmatrix} \Rightarrow c_1 = 0, c_2 = 0, c_3 = 1$

Since A = LDU and Ax = b we have LDUx = b. But we also have Lc = b which implies DUx = c. Substituting from above this gives us

$\begin{bmatrix} d_1&& \\ &d_2& \\ &&d_3 \end{bmatrix} \begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

Multiplying on the left gives us

$\begin{bmatrix} d_1&-d_1&0 \\ 0&d_2&-d_2 \\ 0&0&d_3 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \Rightarrow d_1u-d_1v=0, d_2v-d_2w=0, d_3w=1$

Assuming that none of $d_1$, $d_2$, and $d_3$ is zero, we then have

$d_3w=1 \Rightarrow w = 1/d_3$

$d_2v-d_2w=0 \Rightarrow d_2v = d_2w \Rightarrow v = w = 1/d_3$

$d_1u-d_1v=0 \Rightarrow d_1u = d_1v \Rightarrow u = v = 1/d_3$

So if A is nonsingular then Ax = b has the unique solution

$x = \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 1/d_3 \\ 1/d_3 \\ 1/d_3 \end{bmatrix}$

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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