Projection matrix determinant problem. Is it zero? Please can someone help explain to me why the determinant of all non-identity projection matrices is $0$?
 A: In terms of common sense explanation: a projection matrix projects to a vector subspace by setting the components in the complement of this subspace to zero (unless the subspace = original space and the complement is empty, in which case you have an identity). For instance, projecting onto a plane in 3D sends all vectors along the plane normal to zero. A matrix that sends nonzero vectors to zero is rank-deficient (nonzero kernel) and thus has a determinant 0 (also seen if you notice that sending a subspace to zero corresponds to zero eigenvalues).
A: $A$ is a projection matrix iff $A^2=A$. Hence


*

*The eigenvalues of $A$ belong to $\{0,1\}$.

*$A$ is diagonalisable, since its minimum polynomial can be $p(x)=x,x-1$ or $x^2-x$.
Thus, if $A\ne I$, then it possesses a zero eigenvalue. 
A: Here's an even easier way:
Short version:
$A^2=A, \textrm{det}(A) \ne 0 \Rightarrow A^2 A^{-1} = A A^{-1} \Rightarrow A = I$.
Long version:
By definition, for a projection matrix $A^2=A$. Hence, $\textrm{det}(A)^2 = \textrm{det}(A) \Rightarrow \textrm{det}(A) = 0 \textrm{ or } 1$.
Case 1: zero determinant
Here, the condition that the determinant is zero if the matrix is not the identity is trivially satisfied.
Case 2: unit determinant
Here, the nonzero determinant means that we can invert the matrix, so that
$A^2=A \Rightarrow A^2 A^{-1} = A A^{-1} \Rightarrow A = I$.
Hence, we conclude that if a projection matrix has nonzero determinant it is the identity matrix. Rearranged, this is to say that non-identity projection matrices have zero determinants.
A: Hint  think in terms of eigenvalues $π \lambda_i=det(A)$ so there will always be an eigenvalue when a vector transformed doesnt change and its chatacteristic equation has a root $0$
A: Here’s another way to show this.  
Let $P:V\to W$ be a projection of a finite-dimensional vector space onto a subspace $W$. If $P$ is not the identity, then $\dim W<\dim V$, so the $\dim V$ columns of a matrix that represents $P$, which are the images of a basis for $V$, must be linearly dependent. Therefore, the determinant of the matrix is $0$.
A: Another explanation that doesn't require you to have learned eigenvalues yet:
The projection matrix, $P_V$, for projecting onto a subspace, $V$, of $\mathbb{R}^n$ will be the identity matrix if and only if $V=\mathbb{R}^n$ (since projecting any vector $b \in \mathbb{R}^n$ onto $\mathbb{R}^n$ will yield the same vector). So, if $P_V$ is not the identity matrix, we know that $dim(V) < n$.
If $\{v_1, v_2, ..., v_k\},\space k <n$ is a basis for $V$, then the projection matrix $P_V$ is given by
$$P_V=A(A^TA)^{-1}A^T$$
Where $A$ is the matrix with columns $v_1$ through $v_k$. Since $A$ has fewer than n columns and each column is in $\mathbb{R}^n$, $A$ will be a tall matrix, so it must have at least one zero row in Row Echelon Form. Any matrix that has a zero row in REF has determinant 0. So we have:
$$\det(P_V)=\det(A(A^TA)^{-1}A^T)=\det(A)*\det((A^TA)^{-1}A^T)=0.$$
