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So if I have a random idempotent matrix (a matrix that when squared equals itself) how do I go about calculating its 2-norm?

I know that a idempotent matrix has eigenvalues of only 0 or 1 and I know that in most cases the 2-norm is equal to the largest eigenvalue (although this isn't always the case).

It doesn't make much sense to me.

user1855952
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2 Answers2

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In most cases the 2-norm is equal to the largest eigenvalue.

I don't know what you mean by "most cases"; it is true for normal matrices if you add the absolute value, but not typically otherwise. For nonzero idempotent matrices, the norm is $1$ if and only if the matrix is self-adjoint (if and only if it is normal). In general, the norm of $T$ is the square root of the largest eigenvalue of $T^*T$ (or of $TT^*$).

If $E$ is an $n$-by-$n$ complex idempotent matrix acting on $\mathbb C^n$, and $K\subseteq \mathbb C^n$ is its range, then you can find an orthonormal basis $(e_1,\ldots,e_k)$ for $K$, and complete it to an orthonormal basis $B$ for $\mathbb C^n$; the remaining vectors form an orthonormal basis for $K^\perp$. With respect to this basis, $E$ has matrix

$$[E]_B =\begin{bmatrix}I & A\\ 0& 0\end{bmatrix},$$

for some $k$-by-$(n-k)$ matrix $A$. Note that $A=0$ if and only if $E$ is self-adjoint.

Then the norm of $E$ depends on $A$ according to the equation $\|E\|^2=\|I+AA^*\|=1+\|AA^*\|.$ Since $AA^*$ is positive semidefinite, $\|AA^*\|$ is its largest eigenvalue.

Jonas Meyer
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The 2-norm is square root of the largest eigenvalue of $A^TA$, which is the largest absolute value of an eigenvalue of $A$ only if $A$ is normal. Yours is not.

Here, you might be stuck maximizing: \begin{gather} \|A\|_2^2 = \sup_{\|x\|\neq 0}{\frac{x^TA^TAx}{x^Tx}} \end{gather}

Guido Kanschat
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