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I have encountered a TF question which states that if $A$ is an $n$ by $n$ matrix with $\det(A-3I_n）= 0$， then there is a vector $v$ with $Av = 3v$.

I think about $A$ as a scaling vector, so $A$ should also be $kI_n$, and in this case it would be $3I_n$. Then $\det(A-3I_n)$ becomes $\det(0)$.

My question is that whether $\det(0)$ has a meaning and if so is it equal to zero?

Also, is there any logical fallacy in my approach of this question?

Siong Thye Goh
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user7544590
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  • Your second paragraph does not make sense. There no reason for "$A$ would be 3In". – Mariano Suárez-Álvarez Oct 29 '17 at 18:37
  • See [this question regarding determinants](https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant), it may be of help. – Eff Oct 29 '17 at 18:44
  • Eff, I read this post and by using the properties of determinant I get that det(A) = 3. Does that denote anything other than the matrix A is invertible regarding this question? – user7544590 Oct 29 '17 at 18:49
  • That’s an incorrect conclusion. Determinants are *not* additive. As a counterexample, consider $A=\tiny{\begin{bmatrix}3&0\\0&0\end{bmatrix}}$. It’s obviously singular, but it also satisfies $\det(A-3I)=0$. – amd Oct 30 '17 at 00:03

2 Answers2

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You are correct in asserting that there is a vector $v$ such that $Av = 3v$. Then you go on to think that $Av = 3I_n v$ for every vector $v$, which need not be true.

It is true that the determinant of the zero matrix is $0$, but that's not where your problem lies.

Ethan Bolker
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  • Then is it true to say that A is a scaling vector? Why is it wrong to assume that Av = 3In v in this case? – user7544590 Oct 29 '17 at 18:44
  • Based on your statement of the question, it says that some $v$ always satisfies $Av=3v$. This isn't very interesting because of course such a $v$ always exists, namely the zero vector. If you put a constraint on $v$ being non-zero or stating that any such $v$ should satisfy the claim, then you'd have a more interesting question. – Zachary Oct 29 '17 at 18:54
  • @Zachary, so then if we assume v is non-zero. I am able to derive that det(A) is 3. Does that tell anything about Av = 3v? – user7544590 Oct 29 '17 at 18:56
  • Try using $A=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ and check $det(A-3I)$. Is there then a non-zero $v$ that satisfies $Av=3Iv$? – Zachary Oct 29 '17 at 19:02
  • There is nothing you can say about the determinant of $A$ just knowing that there is one nonzero vector $v$ such that $Av = 3v$. – Ethan Bolker Oct 29 '17 at 19:02
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Example that hopefully helps clear things up.

Consider $A = \begin{bmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3\end{bmatrix}$

Check that $$A\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}= 3\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$

As to how to address the question:

If $\det(A-3I)=0$, what can you say about the rank of $A-3I$? what can you say about the nullity?

Remark: you might to restrict $v$ such that it is non-zero or the question is trivial.

Siong Thye Goh
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