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Find the values of $x$ for which the matrix

\begin{bmatrix} x & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & x \\ \end{bmatrix}

has the lowest rank.

Since the rank is the dimension of the row space of $A$, I should find $x$ for which the dimension of the row space is as low as it can possibly be. I performed Gaussian elimination on $A$ via the following steps:

\begin{bmatrix} x & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & x \\ \end{bmatrix}

\begin{bmatrix} 1 & 0 & \frac{1}{x} \\ 0 & 1 & 0 \\ 1 & 0 & x \\ \end{bmatrix}

\begin{bmatrix} 1 & 0 & \frac{1}{x} \\ 0 & 1 & 0 \\ 0 & 0 & x -\frac{1}{x} \\ \end{bmatrix}

So I should now let $x - \frac{1}{x} = 0 $, more specifically let $x^2 = 1$ and thus $x = \pm 1$. Overall resulting in an overall lowest rank of $2$.

Is this the correct answer? And is it the correct way of thinking about questions such as these? Should I have considered the null space instead? Thanks in advance.

Rodrigo de Azevedo
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zzxxx123
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    That looks right –  May 12 '19 at 14:16
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    For $x = 0$, clearly $\operatorname{rank}(A) = 3$. For $x \neq 0$, what you did is correct. Also, note that if $c_{i}(A)$ ($1 \leq i \leq 3$) denote the columns of $A$, then for any $x \in \mathbb{R}$, $\lbrace c_{2}(A), c_{3}(A) \rbrace$ are two linearly independent vectors. Therefore, $\operatorname{rank}(A) \geq 2$ for all $x$. – pitchounet May 12 '19 at 14:18
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    Your answer is correct. Note that you implicitly assume $x\neq 0$ as you write $\frac{1}{x}$. If $x=0$, the rank of the matrix is $3$ so it doesn't matter. – Mathematician 42 May 12 '19 at 14:18

1 Answers1

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Your solution is correct.

For an alternative, slightly more intuitive solution: notice that no matter the value of $x$,

$$\left\{ \begin{bmatrix}x\\0\\1\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix} \right\} \text{ and } \left\{ \begin{bmatrix}0\\1\\0\end{bmatrix}, \begin{bmatrix}1\\0\\x\end{bmatrix} \right\}$$

will both be linearly independent sets, and so the minimum rank is $2$. To have rank $2$,

$$\left\{ \begin{bmatrix}x\\0\\1\end{bmatrix},\begin{bmatrix}1\\0\\x\end{bmatrix} \right\}$$ must be a linearly dependent set.

Two vectors are linearly independent if and only if one is a multiple of the other. This is clearly only the case here when $x=1$ or $x=-1$, so the matrix has minimal rank when $x = \pm 1$ as you concluded.

To answer your other questions: yes, this is a good way of going about this. Your method is in fact better than mine for the more general case. I just wanted to show you that in this case you can spot the answer without needing to do any row reduction. You could have considered the null space, but by the rank-nullity theorem this is equivalent to what you’ve done, since you would just find that the null space has dimension $1$ implying that the rank is $3-1=2$.

雨が好きな人
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