Similarity of $A$ and $A^{-1}$

Algebra Level pending

Given that $\overrightarrow{\alpha_1},\overrightarrow{\alpha_2},\overrightarrow{\alpha_3}$ are column vectors and they form a basis for $\mathbb R^3$ , $A$ is a $3 \times 3$ matrix such that:

$\begin{cases} A\overrightarrow{\alpha_1}=\overrightarrow{\alpha_1} \\ A\overrightarrow{\alpha_2}=\overrightarrow{\alpha_1}+\overrightarrow{\alpha_2} \\ A\overrightarrow{\alpha_3}=\overrightarrow{\alpha_2}+\overrightarrow{\alpha_3} \end{cases}$

Is it always true that $A$ and $A^{-1}$ are similar , i.e. there exists an invertible matrix $P$ , so that $P^{-1}AP=A^{-1}$ ?

Yes No

1 solution

Karan Chatrath
Jan 4, 2021

We are told that the three vectors ( $v_1$ , $v_2$ and $v_3$ ) are linearly independent. I am not using the vector symbol and and dropping the Greek letter $\alpha$ for the sake of typing convenience. So it is given that:

$Av_1 = v_1$ $Av_2 = v_1+v_2$ $Av_3 = v_2+v_3$

It is obvious from the first equation that $v_1$ is an eigenvector of the matrix $A$ and the corresponding eigenvalue is unity. Looking at the second equation:

$(A-I)v_2 = v_1$ $(A-AA^{-1})v_2 = v_1$ $A(I-A^{-1})v_2 = v_1$ $(I-A^{-1})v_2 = A^{-1}v_1$

$\because Av_1 = v_1 \implies A^{-1}v_1 = v_1$ $(I-A^{-1})v_2 = A^{-1}v_1=Av_1 = v_1$ $(I-A^{-1})v_2=(A-I)v_2 = v_1$ $\implies I-A^{-1}=A-I$ $\boxed{A + A^{-1} =2I}$

Let us look at the third equation now:

$(A-I)v_3 = v_2$

We already know that: $(I-A^{-1})v_2= v_1$ $\implies (I-A^{-1})(A-I)v_3 = v_1$ $\implies (A + A^{-1} -2I)v_3 = v_1$ $\therefore v_1 = 0$

So, $v_1$ is essentially the null vector. Coming back to the relations:

$(I-A^{-1})v_2=(A-I)v_2 = v_1$

If $v_2$ is a nonzero vector, then it must be true that:

$A = A^{-1}=I$

This makes sense as an eigenvalue of the identity matrix is unity and all vectors in $R^3$ are eigenvectors of the identity matrix. So essentially:

$P^{-1}AP = A^{-1} \implies P^{-1}P = I$

Which is an identity for every invertible matrix $P$ . So the answer to the question is $\boxed{\mathrm{YES}}$

To Be honest, I am not convinced by my own solution now that I think more about it. I am open to counter-arguments. Concluding that $v_1$ is a null vector contradicts the given information that the given vectors are linearly independent. Or am I making a more fundamental conceptual error?

Karan Chatrath - 5 months, 1 week ago

In the line $(I-A^{-1})v_2 =(A-I)v_2 =v_1$ , $(I-A^{-1})v_2 =(A-I)v_2$ does not guarantee that $I-A^{-1} = A-I$ .

Alice Smith - 5 months, 1 week ago

Similarity of A A A and A − 1 A^{-1} A − 1

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Similarity of $A$ and $A^{-1}$