## 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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