The matrix of a projection can never be invertible

Question

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?

Answer 1

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 $P=I$.

Answer 2

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.

Answer 3

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 $P=I$.

Answer 4

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

Answer 5

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.