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"Let $V$ be a finite-dimensional vector space and T be the projection on $W$ along $W'$, where $W$ and $W'$ are subspaces of $V$. Find an ordered basis $\beta$ for $V$ such that $[T]_\beta$ is a diagonal matrix."

Playing around in $R^2$ and $R^3$ I found it difficult to reach a diagonal matrix. E.g. let $\beta = \{(1 1 1),(101) \}$ be a basis of $W$ and $\gamma = \{ (001)\}$ be a basis for $W'$. Then $T(abc) = (aba)$. Any basis of $V$ must contain some vector $v_i$ with a value at $a$, e.g. $(100)$ which means that $T(100) = (101)$, which again means that in one column of $[T]_\beta $there will be more than one value, so that it can't be a diagonal matrix.

Now... what did I get wrong?

thanks !

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up vote 0 down vote accepted

Any projection is diagonalizable. In the example you give, $T$ is diagonal in the basis $(1,0,1),(0,1,0),(0,0,1)$.

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thank you. Thanks to your answer I just figured what I thought of wrong. – foaly Nov 14 '12 at 13:59

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