# 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?

• 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. Commented Nov 18, 2012 at 17:29
• Since already answered, let's just add for completeness than any polynomial $p(A)$ is also singular. Commented Nov 18, 2012 at 21:21
• At least if $p$ does not have a constant term. Commented Nov 19, 2012 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.

• This sounds like the plot for a good action movie. Commented Nov 18, 2012 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

• Earned a badge for this answer :) Commented Nov 18, 2012 at 16:57
• This makes so much more sense than the current 'top' answer. Well done for clarity. :) Commented Nov 18, 2012 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. Commented Nov 18, 2012 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 :) Commented Nov 18, 2012 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. :) Commented Nov 18, 2012 at 23:57

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

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.