Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

1 Answer 1

up vote 4 down vote accepted

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

share|improve this answer
    
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.