# Matrix proof $A ^ { - 1 } \mathrm { x } = \frac { 1 } { \lambda } \mathrm { x }$

Prove that, if $$A$$ is an invertible $$n \times n$$ matrix and $$A \mathrm { x } = \lambda \mathrm { x }$$ for some non-zero n-vector $$\mathrm { x }$$ and some scalar $$\lambda ,$$ then

$$A ^ { - 1 } \mathbf { x } = \frac { 1 } { \lambda } \mathbf { x }.$$

So I did simple algebra, but I don't know if the inverse of A can equal the reciprocal of $$\lambda$$ here:

$$A x = \lambda x$$

$$A = \lambda$$

$$A ^ { - 1 } = \lambda ^ { - 1 }$$

$$A ^ { - 1 } = \frac { 1 } { \lambda }$$

$$\therefore A ^ { - 1 } x = \frac { 1 } { \lambda } x$$

• @user376343 Even with that correction, the step in the derivation is still erroneous. You can’t conclude from $Ax=\lambda x$ for some $x$ that $A=\lambda I$.
– amd
Feb 5, 2019 at 21:30
• Right @amd Thinking correctly, writing wrongly. Feb 5, 2019 at 21:34

You can't say $$A=\lambda$$, because $$A$$ is a matrix and $$\lambda$$ is a number. Even if you write the equality as $$A\mathbf{x}=(\lambda I)\mathbf{x}$$ you cannot deduce that $$A=\lambda I$$ ($$I$$ the identity matrix), but just that $$\mathbf{x}$$ belongs to the null space of $$A-\lambda I$$.

First observe that $$\lambda\ne0$$, otherwise $$A\mathbf{x}=\mathbf{0}$$, contradicting $$A$$ being invertible. Then multiply by $$A^{-1}$$, so $$\mathbf{x}=A^{-1}(\lambda\mathbf{x})=\lambda(A^{-1}\mathbf{x})$$ Now multiply by $$\lambda^{-1}$$: $$\lambda^{-1}\mathbf{x}=A^{-1}\mathbf{x}$$

\begin{align} Ax &= \lambda x\\ A^{-1} A x &= A^{-1} \lambda x\\ \frac{1}{\lambda} x &= A^{-1} x \end{align}

First left-multiply both sides by $$A^{-1}$$. Then the $$A$$ and $$A^{-1}$$ cancel on the left side. Then multiply both sides by $$\frac{1}{\lambda}$$ to get your result.

• Omg , its so clear now , thanks to think I raised a matrix to the power -1 face palm Feb 5, 2019 at 15:46

That does not work, you can't take an inverse of a vector. Right now at some point it says a matrix ($$A$$) is equal to a scalar ($$\lambda$$). Rather use that $$x=A^{-1}Ax=\lambda A^{-1}x$$ We can now divide by $$\lambda$$ on both sides to get $$\frac{1}{\lambda}x=A^{-1}x$$.

Multiply $$Ax=\lambda x$$ by $$A^{-1}$$ on both sides, then $$x=\lambda A^{-1}x$$ and, since $$A$$ is invertible, $$\lambda \neq 0.$$ So we can divide by $$\lambda.$$