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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:

$A x = \lambda x$

$A = \lambda$

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

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

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

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  • $\begingroup$ @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 $A=\lambda I$. $\endgroup$
    – amd
    Feb 5, 2019 at 21:30
  • $\begingroup$ Right @amd Thinking correctly, writing wrongly. $\endgroup$
    – user376343
    Feb 5, 2019 at 21:34

4 Answers 4

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

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\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.

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  • $\begingroup$ Omg , its so clear now , thanks to think I raised a matrix to the power -1 face palm $\endgroup$ Feb 5, 2019 at 15:46
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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 $\frac{1}{\lambda}x=A^{-1}x$.

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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 $\lambda.$

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