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Let $\lambda$ be an eigenvalue of $A$. Prove that $\lambda^{-1}$ is an eigenvalue of $A^{-1}$.

My approach:

Suppose $\lambda$ is an eigenvalue of $A$. Then $Ax=\lambda x$ for some $x\neq 0$. Since $A$ is invertible, $Ax=\lambda x \implies A^{-1}Ax=A^{-1}\lambda x$. So:

$$Ix=A^{-1}\lambda x \iff Ix-A^{-1}\lambda x=0\iff x(I-A^{-1}\lambda)=0 $$

Since $x \neq 0$ and $A$ is invertible, $\lambda^{-1}=A^{-1}$ (I'm unsure of this equality). I think this somewhat shows it, this might seem elementary but aren't $\lambda, \lambda^{-1}$ scalars? So basically, if this proof is true $\lambda\cdot\lambda^{-1}=I$ also?

Any other way to show this? Thanks

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    $\begingroup$ You need to assume $A$ is invertible, $\endgroup$ Aug 6 '15 at 19:53
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    $\begingroup$ You just need to show that there exists a nonzero vector $w$ so that $A^{-1}w = \lambda^{-1} w$. From your attempt, it looks like you already have a candidate! $\endgroup$ Aug 6 '15 at 19:54
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    $\begingroup$ You were almost finished early on. We get $x=A^{-1}(\lambda x)=\lambda A^{-1}x$. Divide by $\lambda$ (it is not $0$ since $A$ is invertible). So we have $A^{-1}x=\lambda^{-1}x$. Much of the rest of the work was very wrong, Of course $\lambda^{-1}$ is not $A^{-1}$. $\endgroup$ Aug 6 '15 at 19:55
  • $\begingroup$ $x(I-A^{-1} \lambda)$ is in the wrong order. And no, $\lambda^{-1}$ is not $A^{-1}$: a scalar is not a matrix. $\endgroup$ Aug 6 '15 at 19:55
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Before I continue, it is important to note that $\lambda$ is a scalar and not a matrix. So given that $A$ is invertible, $Ax=\lambda x$, $A$ is invertible, and $\lambda\neq 0$, we have
$$Ax=\lambda x\implies A^{-1}Ax=A^{-1}\lambda x\implies x=\lambda A^{-1}x\implies \frac1\lambda x=A^{-1}x.$$

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Your proof is fine, but you seem to be making things a bit overly complicated.

Proof: Let $x \neq 0$ be such that $Ax = \lambda x$. Since $A$ is invertible, it has a trivial kernel, so $\lambda \neq 0$. We then have $$ Ax = \lambda x \implies\\ A^{-1}Ax = A^{-1}(\lambda x) \implies\\ x = \lambda A^{-1}(x) \implies\\ \frac 1{\lambda} x = A^{-1}x $$ By the above equation, $x$ is an eigenvector of $A^{-1}$ associated with the eigenvalue $1/\lambda$. The conclusion holds.

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(As Robert Israel mentioned in comments, you must assume $A$ is invertible).

There is a more direct approach you can take. As you said $$Ax=\lambda x \implies A^{-1}Ax=A^{-1}\lambda x$$ and then $$A^{-1}Ax=A^{-1}\lambda x \implies Ix = A^{-1}(\lambda x) = \lambda A^{-1}x$$ Now dividing both sides by $\lambda$ yields $$\lambda^{-1}x = A^{-1}x$$ i.e. $\lambda^{-1}$ is an eigenvalue of $A^{-1}$.

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  1. Your worry about equating scalars to matrices is justified. Don't do it!
    If we had $I-A^{-1}\lambda=0$, then multiplying by $\lambda^{-1}$ (also called $\frac1\lambda$) we would get $\lambda^{-1}I-A^{-1}=0$, i.e. $\lambda^{-1}I=A^{-1}$, which are now two matrices.
  2. Matrix multiplication keeps less properties of number multiplication, e.g. the implication
    '$AB=0\implies A=0$ or $B=0$' $\,$ is lost, so we won't get the conclusion $I-A^{-1}\lambda=0$.
  3. All we need is $$A^{-1}x=\lambda^{-1}x$$ for this particular $x$.
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From $Ix=A^{-1}\lambda x$ you have $A^{-1}x=\lambda^{-1}x$, which shows what you want. You are correct that $\lambda^{-1}$ is a scalar, so cannot equal $A^{-1}$. Their products with the vector $x$ are equal, but that does not say they are equal.

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In general we have the equation

$$ \sum_{k=0}^n a_k \mathbf{A}^k = 0. \tag 1 $$

The characteristic polynomial for the matrix $\mathbf{A}$ is then given by

$$ \sum_{k=0}^n a_k x^k = 0. \tag 2 $$

The zeros of this equation are the eigenvalues.

If no eigenvalue is $0$, then $\mathbf{A}^{-1}$ is defined. So we can write

$$ \mathbf{A}^{-n} \sum_{k=0}^n a_k \mathbf{A}^{k} = \sum_{k=0}^n a_k \mathbf{A}^{k-n} = \sum_{k=0}^n a_{n-k} \Big( \mathbf{A}^{-1} \Big)^{k} = 0. \tag 3 $$

The characteristic polynomial for the matrix $\mathbf{A}^{-1}$ is then given by

$$ \sum_{k=0}^n a_{n-k} \big( x' \big) ^k = 0. \tag 4 $$

Then

$$ \Bigg( \frac{1}{x'} \Bigg)^{-n} \sum_{k=0}^n a_{n-k} \big( x' \big) ^k = \sum_{k=0}^n a_{n-k} \Bigg( \frac{1}{x'} \Bigg)^{n-k} = \sum_{k=0}^n a_k \Bigg( \frac{1}{x'} \Bigg)^k = 0. \tag 5 $$

Compare (2) and (5) and we see that the roots satisfy

$$ x' = \frac{1}{x}. \tag 6 $$

Conclusion:

So IF $\lambda$ is an eigenvalue of the invertible matrix $\mathbf{A}$

THEN $\lambda^{-1}$ an eigenvalue of the matrix $\mathbf{A}^{-1}$.

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You're making it too hard. Say $x\ne0$ and $Ax=\lambda x$.
Then $x=A^{-1}Ax=\lambda A^{-1}x$, so $\lambda^{-1}x=A^{-1}x$.

Details: Of course we need to assume here that $A$ is invertible. And then we need to show that it follows that $\lambda\ne0$, which is easy: If $0$ is an eigenvalue of $A$ then $Ay=0$ for some $y\ne0$, showing that $A$ is not invertible.

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