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I am currently learning Linear Algebra and I reached the topic of matrix diagonalization. What I understood that we do with matrix diagonalization is the following (please correct me if I am wrong):

We have a matrix $A$. I take $A$'s eigenvectors and put them in a matrix $S$ (as colums). The product of $A S$ will be $S \Lambda$, where $\Lambda$ is the diagonal matrix with the eigenvalues and $S$ is the same matrix, the one with the eigenvectors. So, $AS=S \Lambda. $

Then, to get just one matrix on the left side we multiplied with $S^{-1}$ and got $A=S \Lambda S^{-1}$.

Why were we allowed to do that? What guarantee do we have that $S$ is invertible? Why is $S$ invertible? Are the eigenvectors always independent, therefore making $S$ always invertible? If not, we shouldn't be allowed to multiply with $S^{-1}$, rigth?

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  • $\begingroup$ It turns out that, for diagonalizable matrices, this matrix $S$ will be invertible. It has to do with the eigenvectors being linearly independent. $\endgroup$ – Dave Nov 9 '18 at 17:29
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If an $n×n$ matrix has rank $n$, it is invertible.

We are assuming we have a basis consisting of eigenvectors. Hence they are linearly independent and the matrix of eigenvectors has rank $n$.

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$A$ has an infinite number of eigenvectors (any nonzero scalar multiple of an eigenvector is also an eigenvector), so you can’t just throw “the” eigenvectors of $A$ into $S$. The eigenvectors that go into $S$ are specifically chosen to be linearly independent, and so form a basis for the space. The essence of diagonalizability is that it is possible to do so. Now, you are guaranteed that eigenvectors with different eigenvalues are linearly independent, but choosing the vectors that go into $S$ when you have repeated eigenvalues (multiplicity $>1$) requires a bit more care.

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