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

There must be something fundamental about matrices that I don't understand. How can I solve a matrix equation without an invertible matrix? How can a constant $\lambda$ equal a matrix?

$$ LHS=\begin{bmatrix} 3 &-2 &4\\ 1 &0 &2\\ -1 &1 &-1 \end{bmatrix}\cdot \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}= \begin{bmatrix} 3x_1-2x_2+4x_3\\ x_1+2x_3\\ -x_1+x_2-x_3 \end{bmatrix} $$ $$ \\\ RHS=\lambda\cdot\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} \lambda x_1\\ \lambda x_2\\ \lambda x_3 \end{bmatrix} $$

No constant can ever make these two sides equal for all $x_1,x_2,x_3$.

And let's say we try to solve it algebraically:

$$ AX=\lambda X \Longleftrightarrow AX-\lambda X=\textbf{0} \Longleftrightarrow (A-\lambda)X=\textbf{0} $$

the $(A-\lambda)$ doesn't make sense as you can't subtract scalars from matrices.

What is going on here?

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    $\begingroup$ $A-\lambda$ does not make sense. $A - (\lambda \times I)$ does. The point is that $\lambda \times X = \lambda \times IX$, so during factorization you don't pull only the $\lambda$ , but the $\lambda I$ out. $\endgroup$ – астон вілла олоф мэллбэрг Sep 6 '18 at 14:48
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    $\begingroup$ You don't search for $\lambda$ such that the both sides are equal for ALL $x_1,x_2,x_3$, but for $\lambda$ such that the both sides are equal for SOME $x_1,x_2,x_3$, not all $0$. $\endgroup$ – Babelfish Sep 6 '18 at 14:50
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    $\begingroup$ @Chisq It is only a minor mistake occasionally made. Always good to ask. $\endgroup$ – астон вілла олоф мэллбэрг Sep 6 '18 at 14:55
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    $\begingroup$ @Chisq If you stay on this site, I promise you your grades will improve. $\endgroup$ – астон вілла олоф мэллбэрг Sep 6 '18 at 15:03
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    $\begingroup$ Also note that unlike on $\mathbb R$, the multiplication on a matrix space isn't commutative. So $X(A-\lambda I)$ is different from $(A-\lambda I)X$. $\endgroup$ – F.Carette Sep 6 '18 at 15:03
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Note that $$AX=\lambda X \Longleftrightarrow AX-\lambda X=\textbf{0} \Longleftrightarrow (A-\lambda I)X=\textbf{0}$$

where $A-\lambda I$ is a matrix which makes your computation make sense.

The formal way to find the eigenvalue $\lambda$ is to solve the characteristic polynomial, $$P(\lambda) = \det (A-\lambda I)=0$$

for $\lambda$ and find the eigenvectors afterwards.

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You have to solve the homogeneous system $(A - \lambda I)X = 0$.

We can do row reduction:

$$A - \lambda I = \begin{bmatrix} 3-\lambda & -2 & 4 \\ 1 & -\lambda & 2 \\ -1 & 1 & -1-\lambda\end{bmatrix} \sim \begin{bmatrix} 1 & -\lambda & 2 \\ 0 & -(\lambda-1)(\lambda - 2) & 2(\lambda-1) \\ 0 & 1-\lambda & 1-\lambda\end{bmatrix}$$

Now, if $\lambda = 1$ we have

$$A - I \sim \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

so the system has a nontrivial solution $X \in \operatorname{span}\left\{\begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ -1\end{bmatrix}\right\}$.

Assume $\lambda \ne 1$ and continue by dividing the last two rows by $\lambda - 1$:

$$A - \lambda I =\begin{bmatrix} 1 & -\lambda & 2 \\ 0 & \lambda-2 & -2 \\ 0 & 1 & 1\end{bmatrix} \sim \begin{bmatrix} 1 & -\lambda & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$

If $\lambda = 0$, we have

$$A \sim \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

so the system has a nontrivial solution $X \in \operatorname{span}\left\{\begin{bmatrix} 2 \\ 1 \\ -1\end{bmatrix}\right\}$.

If $\lambda \ne 0, 1$, we can divide the last row by $\lambda$ to obtain

$$A - \lambda I \sim \begin{bmatrix} 1 & -\lambda & 2 \\ 0 & 1 & 1 \\ 0 & 0 & \lambda\end{bmatrix}$$

which has only the trivial solution.

We conclude that $(A - \lambda I) X = 0$ has nontrivial solutions if and only if $\lambda \in \{0,1\}$.

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