# Prove Frobenius norm of projection matrix is 1

Given a projection matrix $P$, i.e. $P^2=P$, how do you prove $\|P\|_F=1$, where $\|\cdot\|_F$ is the Frobenius norm of the matrix?

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Is the Frobenius norm defined as $\sqrt{\sum_{i,j=1}|p_{i,j}^2|}$? Is this case, what about $P=I$? – Davide Giraudo Sep 7 '11 at 10:45
By submultiplicativity you get $\|P\|_F\geq 1$, but in general, equality does not hold. – Dirk Sep 7 '11 at 11:32
If you want to know more about the Frobenius norm of orthogonal projections, consider the formula $\|P\|_F = \sqrt{tr(P^T P)}$. For orthogonal projection you can use $P^T=P$ and $P^2=P$ and knowledge about the eigenvalues of projections to obtain that the Frobenius norm is equal to the square root of the dimension of the range of $P$. – Dirk Sep 7 '11 at 11:40

$\|P\|_F\ne 1$ in general.

For orthogonal projections, $P$ is positive semi-definite. The eigenvalue (also singular values) $P$ is either 1 or 0. In $\mathbb{R}^n$, if $P$ projects a vector on to a $m$-dimensional subspace, $P$ will have $m$ eigenvalues as 1.

So $\|P\|_F^2=\sum \lambda_i^2=m$. So $\|P\|_F=1$ only if $m=1$, i.e., $P$ project a vector to a given vector.

If the given vector is $v$, then $$P=\frac{vv^T}{v^Tv}$$ In general, if $P$ project a vector to the column space of a matrix $A$ with full column rank, then $$P=A(A^TA)^{-1}A^T$$

See here for a reference.

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@user1551: Do you mean oblique projections? My answer holds for orthogonal projections. – Shiyu Sep 7 '11 at 14:44
@user1551: it's a typo, obviously it should be positive semi-definite. thanks anyway. – Shiyu Sep 8 '11 at 1:59
@user1551: If $P$ is not Hermitian, its Frobenius norm is always greater than $1$, even if $P$ has rank $1$. – Jonas Meyer Sep 8 '11 at 3:56