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$A$ is the linear mapping $f(x)= Ax,\mathbb{R} \rightarrow \mathbb{R}$

$$f\left( \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix}\right)= \begin{pmatrix} x_{2}-x_{3}\\ x_{1}+3x_{2}-2x_{3}\\ x_{1}-4x_{2}+5x_{3} \end{pmatrix}$$

Is zero an eigenvalue of $A$?

I need the matrix $A$. I think we can get it by forming the thing above? So

$$A= \begin{pmatrix} 0 & 1 & -1\\ 1 & 3 & -2\\ 1 & -4 & 5 \end{pmatrix}$$

I hope this is correct, if not everything else coming now will be wrong :s

$$\begin{vmatrix} -\lambda & 1 & -1\\ 1 & 3-\lambda & -2\\ 1 & -4 & 5-\lambda \end{vmatrix} \begin{matrix} -\lambda & 1 \\ 1 & 3-\lambda \\ 1 & -4 \end{matrix}$$

Characteristic polynomial $p_{A}(\lambda)= -\lambda(3-\lambda)(5-\lambda)+(-2)+4-(-(3-\lambda))-(-8\lambda)-(5-\lambda)$

$p_{A}(\lambda)= (-3\lambda+\lambda^{2})(5-\lambda)+2-(-3+\lambda)+8\lambda-5+\lambda$

$p_{A}(\lambda)= -15\lambda +3\lambda^{2}+5\lambda^{2}-\lambda^{3}+2+3-\lambda+8\lambda-5+\lambda$

$p_{A}(\lambda)= -\lambda^{3}+8\lambda^{2}-7\lambda$

We quickly see that $\lambda=0$ so zero is really an eigenvalue of $A$?

Is it correct? I especially wasn't sure at the beginning if I formed it to a matrix correctly because I didn't work with these.. mappings? in combination with matrix before.

cnmesr
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    By definition of eigenvalue, $0$ is one of $f$ if there exists a non-null vector $v$ such that $f( v)=0\cdot v$. This is equivalent to there existing a non-null vector $v$ such that $f(v)=0_{\mathbb R^3}$. One can even eyeball such a $v$. I didn't check your work, but yes, $0$ is an eigenvalue of $f$. – Git Gud Mar 13 '17 at 16:19
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    Remember that the product of a matrix’s eigenvalues is equal to its determinant, so every singular matrix has $0$ as an eigenvalue. – amd Mar 13 '17 at 18:14
  • Null vector appears in https://math.stackexchange.com/questions/2185048/did-i-do-it-correctly-whats-the-image-and-the-basis-of-the-image-of-this-lineahttps://math.stackexchange.com/questions/2185048/did-i-do-it-correctly-whats-the-image-and-the-basis-of-the-image-of-this-linea – dantopa Mar 13 '17 at 21:25

3 Answers3

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Yes, given your matrix $A$

$$\left( \begin{array}{ccc} 0 & 1 & -1\\ 1 & 3 & -2\\ 1 & -4 & 5\\ \end{array} \right)$$

by inspection the vector $$\left( \begin{array}{c} -1 \\ 1\\ 1\\ \end{array} \right)$$

is an eigenvector of $A$ with the required eigenvalue you are uncertain about $(0)$

PM.
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It may seem unusual that zero is an eigenvalue for your matrix, however it is allowed. If you look at your characteristic polynomial $ - \lambda ^3 + 8 \lambda ^2 - 7 \lambda = -\lambda ( \lambda ^2 + 8 \lambda - 7 ) $, you see that zero is indeed a root for this polynomial - you just have to find the corresponding eigenvector.

There are certain matrices actually that only have zero as an eigenvalue - namely nilpotent matrices, it may be worth having a look at the proof: Prove that the only eigenvalue of a nilpotent operator is 0?

Community
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Mathew Duxbury
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Observe that the operation $C_1\rightarrow C_1-C_2-C_3$ produces a zero column in first place hence det($A$)$=0$. So $0$ must be an eigenvalue.

Nitin Uniyal
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