## Linear Algebra and Its Applications, Exercise 2.3.3

Exercise 2.3.3. Given the general triangular matrix

$T = \begin{bmatrix} a&b&c \\ 0&d&e \\ 0&0&f \end{bmatrix}$

show that the rows of $T$ are linearly dependent if any of the diagonal entries $a$, $d$, or $f$ is zero.

Answer: We can test for linear independence of the rows by doing elimination on a second matrix $A$ whose columns are the rows of $T$:

$A = \begin{bmatrix} 0&0&a \\ 0&d&b \\ f&e&c \end{bmatrix}$

We start by doing a row exchange to exchange the first row with the third:

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

If $a = 0$ we have the following matrix:

$\begin{bmatrix} f&e&c \\ 0&d&b \\ 0&0&0 \end{bmatrix}$

This matrix has at most only two pivots, and might have only one or even zero depending on the values of the other entries, so that the rank $r < n$. So if $a = 0$ the  columns of $A$, and thus the rows of $T$, are linearly dependent.

if $d = 0$ we have the following matrix

$\begin{bmatrix} f&e&c \\ 0&0&b \\ 0&0&a \end{bmatrix}$

Elimination would produce a matrix with two pivots at most, so if $d = 0$ we again have $r < n$ and the columns of $A$, and thus the rows of $T$, are linearly dependent.

Finally if $f = 0$ we have the following matrix

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

Again elimination would produce a matrix with two pivots at most, so if $f = 0$ we again have $r < n$ and the columns of $A$, and thus the rows of $T$, are linearly dependent.

We conclude that if any of $a$, $d$, or $f$ are zero then the rows of $T$ are guaranteed to be linearly dependent.

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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