Linear Algebra and Its Applications, Review Exercise 2.31

Review exercise 2.31. Consider the rank-one matrix $A = uv^T$. Under what conditions would $A^2 = 0$?

Answer: In order for $A^2$ to exist $A$ must be a square matrix; otherwise we could not multiply $A$ by $A$ since the number of columns of the first matrix would not match the number of rows of the second.

Since $A$ is an $n$ by $n$ matrix the vectors $u$ and $v$ both have $n$ entries, with $u = (u_1, \ldots, u_n)$ and $v = (v_1, \ldots, v_n)$.

We then have

$A^2 = (uv^T)^2 = uv^Tuv^T = u(v^Tu)v^T$

Since $v^T$ is 1 by $n$ and $u$ is $n$ by 1 their product $v^Tu$ is a 1 by 1 matrix, or in other words a scalar value

$c = \sum_{i=1}^n v_iu_i = \sum_{i=1}^n u_iv_i$

We thus have

$A^2 = u(v^Tu)v^T = ucv^T = cuv^T = cA$

One way for $A^2$ to be zero is to have $A = 0$. Another way is if $c = \sum_{i=1}^n u_iv_i = 0$. Note that this would be trivially true if $A = 0$ but it is also possible for $c$ to be zero even if $A$ is nonzero.

For example, consider the case where $u = (1, 1)$ and $v = (1, -1)$ so that

$A = uv^T = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \begin{bmatrix} 1&-1 \end{bmatrix} = \begin{bmatrix} 1&-1 \\ 1&-1 \end{bmatrix}$

We then have

$c = \sum_{i=1}^n u_iv_i = 1 \cdot 1 + 1 \cdot (-1) = 1 - 1 = 0$

and

$A^2 = \begin{bmatrix} 1&-1 \\ 1&-1 \end{bmatrix} \begin{bmatrix} 1&-1 \\ 1&-1 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} = 0$

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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