I have two $n \times n$ matrices A and B. They have same eigenvectors but not necessarily same eigenvalues and both of them are diagonalizable. I want to show that for any real numbers c and d, $E=cA+dB$ is also diagonalizable.

My question is:

Even though they have same eigenvectors, does not the order in which the eigenvectors are put determine what diagonalizing matrix they have? Or in other words, do they need to have same diagonalizing matrix? How do I approach this problem?


Let $V$ be a matrix containing the eigenvectors (of which there are $n$, where these are $n \times n$ matrices). Then $$ V^{-1} A V = D_1 $$ is diagonal, as is $V^{-1} B V = D_2$. (It'd be a really good exercise for you to figure out why,) If you now compute, you get $$ V^{-1}(cA + dB)V = (cV^{-1}A + dV^{-1}B) V = c V^{-1}AV + dV^{-1}BV = cD_1 + dD_2 $$ which is again diagonal. $$D_{(i,i)} = cD_{1(i,i)} + dD_{2(i,i)}$$

  • $\begingroup$ The way I think why it should be true is because even though the two matrices have different eigen values, I can arrange them in such a manner that the corresponding diagnonalizing matrix will be same for both of them, Is that true? $\endgroup$ – Avery Dec 18 '16 at 2:15
  • $\begingroup$ Or I should not worry about the orders in which the eigen vectors are arranged in diagonalizing matrix, because I can easily change the position of eigen values in the diagonal matrices D1 and D2. $\endgroup$ – Avery Dec 18 '16 at 2:16
  • $\begingroup$ Once you pick the eigenvectors and put them in $V$, the corresponding eigenvalues will show up, in $V^{-1}AV$ and $V^{-1}BV$, at the corresponding places. You really should work out that exercise I mentioned in the middle...if you actually want to understand, I mean. $\endgroup$ – John Hughes Dec 18 '16 at 2:42
  • $\begingroup$ Oh yeah. I understand that part. My only confusion was why V should be the same for both A and B. and I think you have made it clear. $\endgroup$ – Avery Dec 18 '16 at 2:44

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