# why isn't $cI_2$ diagonalizable

A question in my linear algebra textbook asks me to prove that $cI_2$ is not diagonalizable. Since an $n\times n$ matrix $A$ is diagonalizable only if it has $n$ linearly independent eigenvectors, we know that $cI_2$ is not diagonalizable. However, using a different approach, I reached the opposite conclusion and can't figure out where I went wrong.

By definition, a matrix $A$ is diagonalizable if $P^{-1}AP = D$, and $D$ is diagonal. So,

\begin{align} D & = P^{-1}AP \\ & = P^{-1}(cI_2)P \\ & = c(P^{-1}IP)\\ & = c(P^{-1}P)\\ & = cI \end{align}

Since $cI$ is diagonal, $D$ is diagonal, and $cI$ is therefore diagonalizable. A similar approach I thought of is to let $P=I$ so that

\begin{align} D & = P^{-1}(cI_2)P \\ & = c(I^{-1}II)\\ & = cI^3\\ & = cI \end{align}

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Are you sure the notation $I_2$ refers to an identity matrix? – Hurkyl Nov 30 '12 at 7:45
@Hurkyl I believe the convention is that $I_2$ refers to the $2\times2$ identity matrix. However, in the book they actually just draw the matrix. – Matt Munson Nov 30 '12 at 8:50
@MattMunson What kinda of textbook is this? Throw it out! – user38268 Nov 30 '12 at 11:33

## 1 Answer

You are correct. I is diagonalizable, and all diagonal matrices are (almost) by definition diagonalizable since one may choose (as you do), $P=I$.

(I am currently teaching in a Linear algebra course).

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Ok, awesome. It seems the text mistakenly concludes that $cI$ does not have $n$ linearly independent eigenvectors. Thanks. – Matt Munson Nov 30 '12 at 6:43
Ah, well ,there's your error then, since clearly, EVERY vector is an eigenvector of cI. – Per Alexandersson Nov 30 '12 at 8:04