Is there a way to interpret the determinant of 0?

Question

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?

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, 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

score 1 · Accepted Answer · answered Oct 29 '17 at 18:37

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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.

answered Oct 29 '17 at 18:37

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

score 0 · Answer 2 · answered Oct 29 '17 at 18:57

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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.

answered Oct 29 '17 at 18:57

Siong Thye Goh

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I only know that the rank of a matrix is n if the determinant is not 0. – user7544590 Oct 29 '17 at 19:02
If the determinant is $0$, the matrix is singular, so can you compare rank of the matrix with $n$? – Siong Thye Goh Oct 29 '17 at 19:03
so the rank would be 1? and the nullity would be 3-1 = 2 in this case – user7544590 Oct 29 '17 at 19:04
In my particular example, the rank is $1$, in general we can only say that rank is less than $n$ and hence the nullity is positive. Think of what does nullity is positive means. =) – Siong Thye Goh Oct 29 '17 at 19:08

Is there a way to interpret the determinant of 0?

2 Answers2