Given $A \in \Bbb C^{3 \times 3}$ that has a SVD $A=u_1v_1^*+u_2v_2^*+u_3v_3^*$ then is $A$ a unitary matrix?

Question

Given $A \in \Bbb C^{3 \times 3}$ that has a SVD $A=u_1v_1^*+u_2v_2^*+u_3v_3^*$ then is $A$ a unitary matrix?

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We know that every matrix $A \in \Bbb C^{m \times n}$ has a SVD. Some things that I am not sure of are:

1. Does it mean that if a matrix has SVD then it is Hermitian? because we need $A^* A$ and all eigenvalues are positive?

2. If it is Hermitian then does it mean that it has an orthonormal basis?

3. SVD and spectral theorem seem similar Are they actually connected? What's the difference?

My approach is (assuming that the points above are correct)

• since $A$ has SVD we get that ${u_1,u_2,u_3}$ and ${v_1,v_2,v_3}$ are orthonormal basis

• now we calculate $AA^*$ and get $(u_1 v_1^*+u_2 v_2^* + u_3 v_3^*)(u_1 v_1^*+u_2 v_2^* + u_3 v_3^*)^*$

But I got stuck here. The answer in the textbook is just

this is a spectral decomposition of the unit matrix $I$ therefore $A^*A=I$ and $A$ is unitary matrix

However, I could not really understand how they got to all that information. Is my way correct? How would I continue from there?

Answer 1

Every matrix has an SVD regardless of whether it's Hermitian.

If $A=u_1v_1^*+u_2v_2^*+u_3v_3^*\triangleq U\Sigma V^*$, where $u_i$ and $v_i$ are the $i$th column vectors of unitary matrices $U$ and $V$ respectively, then $$A = UV^*$$ Since $A$ is square, $\Sigma=I$ a unitary matrix. Therefore, $A$ is the product of three unitary matrices and $A$ is unitary.

Addendum:

Spectral decomposition or eigendecomposition is done on square matrices whose eigenspaces span the vector space. SVD is applicable to all matrices. The unitary matrices of SVD have the nice property that $U^*=U^{-1}$ and $V^*=V^{-1}$ (in analogy to eigendecomposition)