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Assuming $D$ is a square matrix where $D^2 = D$ why is $D$ not equal to the Identity matrix.

Assuming $I$ is the identity matrix then why is the following not true.

$D^2 = D$

$DD = D$

$D^{-1}DD =D^{-1}D$

$ID = I$

$D = I$

As far as I can see, this should be true. But if D = $\begin{bmatrix}\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}\end{bmatrix}$. Then $D^2=D$ still holds true yet D is not the identity matrix.

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    Why is this a problem? Or rather: why do you think this *should* be true? – 2017-02-07
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    You multiply both sides by $D^{-1}$. Are you **certain** $D^{-1}$ always exists? – 2017-02-07
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    You are assuming D is invertible – 2017-02-07
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    It *is* true that an invertible matrix $D$ such that $D^2=D$ is equal to the identity matrix (In fact, one can show that a matrix $D$ such that $D^2=D$ is invertible if and only if it is equal to the identity matrix) Your example matrix, on the other hand, is not invertible. – 2017-02-07

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Matrices, that fulfill $D^2=D$ are projections. There is only one projection, that maps into the whole vector-space. It's the identity. Your example projects everything on the space spanned by $\begin{pmatrix}1\\1\end{pmatrix}$ and is not invertible.

In your derivation you asusmed the matrix to be non-singular, but is (basically) never the case in projections.