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I am currently studying linear transformations in order to refresh my knowledge of linear algebra. One statement in my textbook (by David Poole) is:

When considering linear transformations from $\mathbb{R}^2$ to $\mathbb{R}^2$, the matrix of a projection can never be invertible.

I know that a projection matrix satisfies the equation $P^2 = P$. Taking determinant of both sides gives

$$\text{det}(P)^2 = \text{det}(P)$$

which is always true when $P$ is singular. However take $\color{blue} {P = I_2}$, then the equality is true and the projection matrix is invertible. What mistake do I make in my reasoning?

Rodrigo de Azevedo
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5 Answers5

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Well, the statement is plainly false when $P=I$. However, the only invertible projection matrix is the identity. To see this, notice that $P^2x=Px$ for all $x$. So if $P$ were invertible, we get $Px=x$ for all $x$, and since the identity is unique, we get $P=I$.

Alex R.
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Technically this is not true. The only invertible projection is the identity. Proof:

If $P$ is invertible, then $P^{-1}(P^2)=P^{-1}P$ so $P=I$.

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You can find the matrix associated with the the transformation projection. The matrix is singular which implies the projection is not invertible.

Simran Kaur
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A projection (non-trivial one, that is) is not injective. For instance, any two vectors perpendicular to the subspace you are projecting onto go to zero.

suckling pig
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This isn't entirely true, you can have the identity, but assume it's not the identity, then the minimal polynomial divides $x^2-x$ and is not just $x-1$, hence $x$ divides the minimal polynomial. But then as $0$ is a root of the polynomial, it is an eigenvalue for the matrix, $P$, hence $P$ cannot be invertible as its determinant is the product of its eigenvalues.

Adam Hughes
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    What you wrote is right but seems completely overkill for proving such a basic fact. –  Jan 06 '17 at 22:41
  • @ZacharySelk I'm not sure I see your point, there's nothing beyond the scope of a first course here aside from maybe polynomial division which is from high school. – Adam Hughes Jan 06 '17 at 22:46
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    I don't know about you, but minimal polynomials never came up in my undergrad linear algebra course. It was mostly tedious matrix computations. That's besides the point though. You can prove it directly from the definition. I'm not saying anything good or bad, just a comment. –  Jan 06 '17 at 22:49
  • @ZacharySelk of course, I didn't take it as an insult it's just strange to hear. I suppose there may be some variation, so it's probably particularly good to have multiple approaches, especially as projections are not basic topics from primarily computation-based classes. – Adam Hughes Jan 06 '17 at 22:51
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    You're right, projections probably won't be covered in a computation based class. Your answer is helpful, different approaches are always good. +1 –  Jan 06 '17 at 22:54