## Linear Algebra and Its Applications, Review Exercise 2.13

Review exercise 2.13. For the matrix

$A = \begin{bmatrix} a&a&a&a \\ a&b&b&b \\ a&b&c&c \\ a&b&c&d \end{bmatrix}$

find its triangular factors $A = LU$ and describe the conditions under which the columns of $A$ are linearly independent.

Answer: We start elimination by subtracting 1 times the first row from the second row, with $l_{21} = 1$

$\begin{bmatrix} a&a&a&a \\ a&b&b&b \\ a&b&c&c \\ a&b&c&d \end{bmatrix} \Rightarrow \begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ a&b&c&c \\ a&b&c&d \end{bmatrix}$

$L = \begin{bmatrix} 1&&& \\ 1&1&& \\ ?&?&1& \\ ?&?&?&1 \end{bmatrix}$

We next subtract 1 times the first row from the third row, with $l_{31} = 1$

$\begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ a&b&c&c \\ a&b&c&d \end{bmatrix}\Rightarrow \begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&b-a&c-a&c-a \\ a&b&c&d \end{bmatrix}$

$L = \begin{bmatrix} 1&&& \\ 1&1&& \\ 1&?&1& \\ ?&?&?&1 \end{bmatrix}$

and then subtract 1 times the first row from the fourth row, with $l_{41} = 1$

$\begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&b-a&c-a&c-a \\ a&b&c&d \end{bmatrix} \Rightarrow \begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&b-a&c-a&c-a \\ 0&b-a&c-a&d-a \end{bmatrix}$

$L = \begin{bmatrix} 1&&& \\ 1&1&& \\ 1&?&1& \\ 1&?&?&1 \end{bmatrix}$

Turnng to the second column, we subtract 1 times the second row from the third row, with $l_{32} = 1$

$\begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&b-a&c-a&c-a \\ 0&b-a&c-a&d-a \end{bmatrix} \Rightarrow \begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&0&c-b&c-b \\ 0&b-a&c-a&d-a \end{bmatrix}$

$L = \begin{bmatrix} 1&&& \\ 1&1&& \\ 1&1&1& \\ 1&?&?&1 \end{bmatrix}$

and then subtract 1 times the second row from the fourth row, with $l_{42} = 1$

$\begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&0&c-b&c-b \\ 0&b-a&c-a&d-a \end{bmatrix} \Rightarrow \begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&0&c-b&c-b \\ 0&0&c-b&d-b \end{bmatrix}$

$L = \begin{bmatrix} 1&&& \\ 1&1&& \\ 1&1&1& \\ 1&1&?&1 \end{bmatrix}$

Finally we subtract 1 times the third row from the fourth row, with $l_{43} = 1$

$\begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&0&c-b&c-b \\ 0&0&c-b&d-b \end{bmatrix} \Rightarrow \begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&0&c-b&c-b \\ 0&0&0&d-c \end{bmatrix}$

$L = \begin{bmatrix} 1&&& \\ 1&1&& \\ 1&1&1& \\ 1&1&1&1 \end{bmatrix}$

We thus have

$A = LU = \begin{bmatrix} 1&&& \\ 1&1&& \\ 1&1&1& \\ 1&1&1&1 \end{bmatrix} \begin{bmatrix} a&a&a&a \\ 0&b-a&b-a&b-a \\ 0&0&c-b&c-b \\ 0&0&0&d-c \end{bmatrix}$

In order for the columns of $U$ and thus the columns of $A$ to be linearly independent, the values in the four pivot positions must be nonzero, so we must have $a \ne 0$, $b \ne a$, $c \ne b$, and $d \ne c$.

Note that this does not mean that all four values must be nonzero, or that all fur values have to be unique. For example, the conditions would be satisfied if $a = c = 1$ and $b = d = 0$ so that

$A = \begin{bmatrix} 1&1&1&1 \\ 1&0&0&0 \\ 1&0&1&1 \\ 1&0&1&0 \end{bmatrix}$

$= \begin{bmatrix} 1&&& \\ 1&1&& \\ 1&1&1& \\ 1&1&1&1 \end{bmatrix} \begin{bmatrix} 1&1&1&1 \\ 0&-1&-1&-1 \\ 0&0&1&1 \\ 0&0&0&-1 \end{bmatrix} = LU$

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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