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$A$ is a $7 \times 7$ matrix with three eigenvalues. One eigenspace is two-dimensional, and one of the other eigenspaces is three-dimensional. Is it possible that $A$ is not diagonalizable? Justify your answer.

cannot be diagonalizable.

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 3

Diagonalization

Vectors

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Okay, so we have a matrix A which has three again values all in the one and two and three and the first Eggen Spaces of Dimension to and the second Eggen spaces of that mentioned three. Or this sir, we only know that there's one for sure with dimension to one for sure Would I mentioned three? I took the liberty of labeling them first at second. It's not really important. And we wonder, is it possible that this matrix is not dying on advisable? And the answer is yes. It is possible. It is possible that a isn't dia gone uh, realizable. So why is why is it But why is it possible that it isn't diagnosable? Because if the let's say that the third again space is of dimension to then we would have that dimension of the first in space close the dimension of the second Dragon space was that I mentioned of the third Dragon space. Okay, is two plus three plus two, which is seven a is a seven by seven matrix and we have seven literally independent again vectors. Therefore it is diagonal Izabal. So if we have this, then a is Ah diagonal sizable. But let's say we have the something different. Let's say the dimension of the third hag in space is only one. So for the third hag in value, we only get it. I mentioned one for the high in space. So in that case is the dimension of the first haggans spay. There's the dimension of the second hag in space. Trusted, I mentioned of the third haggans space is two plus three plus one. Then we get six For the total I mentioned. That means we only have that we only have Ah, six. Then we're the independent hag in Victor's and a is seven by seven. Therefore, because six six is not equals two Selves six isn't equal to seven. Pretty obvious. A isn't Diagon Izabal. So, yes, it is possible that this maitresse is not diagonal Izabal. If the dimension of the remaining haggans spaces only one then our matrix isn't I noted Sizable

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