# If $A$ is singular, is $A^3+A^2+A$ singular?

Suppose that $A$ is singular, is $A^3 + A^2 + A$ singular as well?

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I would think of that this way: $A$ is a linear transformation from $\mathbb R^n$ to $\mathbb R^n$ that squashes at least one dimension. If you apply this transformation multiple times, it will still squash that dimension. Therefore, that sum is also singular. –  Martin Ueding Nov 18 '12 at 17:29
Since already answered, let's just add for completeness than any polynomial $p(A)$ is also singular. –  peterph Nov 18 '12 at 21:21
At least if $p$ does not have a constant term. –  Gregor Bruns Nov 19 '12 at 0:55

Since $A$ is singular, it has a non-trivial kernel. Let $v$ be a non-zero vector killed by $A$.

Show that $A^3+A^2+A$ kills $v$ too.

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This sounds like the plot for a good action movie. –  Zolomon Nov 18 '12 at 14:00

$A$ is singular hence $|A|=0$ thus: $$|A^{3}+A^{2}+A|=|A(A^{2}+A+I)|=|A||A^{2}+A+I|=0\cdot|A^{2}+A+I|=0$$

hence $A^{3}+A^{2}+A$ is also singular

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Earned a badge for this answer :) –  Belgi Nov 18 '12 at 16:57
This makes so much more sense than the current 'top' answer. Well done for clarity. :) –  Noldorin Nov 18 '12 at 18:07
@Noldorin. I do not agree. This is correct and easy of course, but relies on a few theorems (Singular = zero determinant, multpilicity of the determinant). The other solution requires nothing. –  the L Nov 18 '12 at 18:40
Thank you Noldorin, I really like Mariano answer and as well. as anonymous said it is more basic, and is quite easy as well :) –  Belgi Nov 18 '12 at 18:41
@Belgi: Indeed, there is nothing wrong with Mariano's answer. But it relies on terminology like "kernel" (not usually covered in introductory linear algebra courses) and "kill a vector" (never seen this). Maybe just a British/European thing though. In any case, I think the more steps the better, especially for more 'basic' questions. :) –  Noldorin Nov 18 '12 at 23:57

If $A$ is singular, then $AB$ is singular too, for any choice of $B$. Just put $B=A^2+A+I$.

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+1 for the answer, $+I$ for the edit... :-) –  Amihai Zivan Nov 18 '12 at 16:18
@Amihai: edited :-) –  TonyK Nov 18 '12 at 19:28

A square matrix is singular if and only if there exists a nonzero vector that maps to zero when transformed by the matrix.

Let $A\in\mathbb{R}^\mathit{n{\times}n}$ be singular. Then there exists $x\in\mathbb{R}^n$ such that $x\neq0$ and $Ax=0$. Thus $(A^3+A^2+A)x=(A^2+A+I)Ax=(A^2+A+I)0=0$. Therefore $A^3+A^2+A$ is singular.

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