## 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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