## 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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