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I have found an exercise involving a $3\times 3$ projection matrix which projects to a space $V$ with $\mathrm{dim}(V)=2$.
The matrix(or operator) is defined as $P(x)=v*(x, v)+u*(x, u)$.
So, in my understanding it projects vectors $x$ onto two vectors $u$ and $v$.
Now, the exercise asked us to find it's eigenvalues. So, I found three of them and each one belongs to a different eigenvector(the set of the eigenvectors is orthogonal).
So, upon solving the exercise, I found myself really not understanding this conceptually.

Why would there be 3 orthogonal eigenvectors which span a space of dimension equal to 3 when the projector projects to a space of dimension equal to 2?

EDIT: This is the exercise:
$\def\Bra#1{\left\langle#1\right|}\def\Ket#1{\left|#1\right\rangle}\def\Braket#1{\mathinner{\left\langle{#1}\right\rangle}}$ We have an orthonormal basis $\{\Ket{1},\Ket{2},\Ket{3}\}$. We also have the vectors $\Ket{\Psi}=\Ket{1}-\Ket{2}+\Ket{3}$ and $\Ket{\Phi}=\Ket{1}+\Ket{2}$.
We define an operator(matrix) as: $Q=\Ket{\Psi}\Bra{\Psi}+\Ket{\Phi}\Bra{\Phi}$ with $\Bra{\Psi}$ and $\Bra{\Phi}$ being the complex transpose of each vector.

Solution: i find $\Ket{\Psi}=(1, -1 , 1)^T$ and $\Ket{\Phi}=(1, 1, 0)^T$ with inner products $\Braket{\Psi\mid\Psi}=3$, $\Braket{\Phi\mid\Phi}=2$.
To find the eigenvalues of $Q$ i just act with $Q$ of $\Ket{\Psi}$: $Q\Ket{\Psi}=\Ket{\Psi}\Braket{\Psi\mid\Psi}+\Ket{\Phi}\Braket{\Phi\mid\Psi}=3\Ket{\Psi}$ because $\Ket{\Psi}$ and $\Ket{\Phi}$ are orthogonal.
Also i act with $Q$ on $\Ket{\Phi}$: $Q\Ket{\Phi}=\Ket{\Psi}\Braket{\Psi\mid\Phi}+\Ket{\Phi}\Braket{\Phi\mid\Phi}=2\Ket{\Phi}$.
So we have two eigenvalues: $\lambda=2$ with eigenvector $\Ket{\Phi}$ and $\lambda=3$ with eigenvector $\Ket{\Psi}$.
The solution says that there must be another vector orthogonal to $\Ket{\Phi}$ and $\Ket{\Psi}$ because it is a three dimensional space. That eigenvalue is zero with eigenvector(i have computed it) $(-1,1,2)^T$.

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  • $\begingroup$ Are you sure that all three eigenvectors are associated with the same eigenvalue? $\endgroup$ – amd Apr 30 '16 at 16:33
  • $\begingroup$ @amd "I found three of them and each one belongs to a different eigenvector". No, each eigenvalue belongs to its own eigenvector. $\endgroup$ – TheQuantumMan Apr 30 '16 at 17:16
  • $\begingroup$ Any projection has only two eigenvalues. If you found a third, you did something wrong. $\endgroup$ – amd May 1 '16 at 6:10
  • $\begingroup$ @amd no, the solution of the exercise gives three eigenvalues. It's a 3x3 matrix. Wouldn't we expect to find 3? $\endgroup$ – TheQuantumMan May 1 '16 at 9:49
  • $\begingroup$ I’d expect at most three. For any projection, the only eigenvalues are $0$ and $1$, with multiplicities $>1$ for a $3\times 3$ or larger matrix. $\endgroup$ – amd May 2 '16 at 6:06
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Any non-degenerate projection $P$ onto a (sub)space $V$ has only two eigenvalues: $1$, with associated eigenspace $V$ itself, and $0$, with associated eigenspace $\ker(P)$. In this case, $P$ is an orthogonal projection, so it should come as no surprise that eigenvectors associated with $0$ are orthogonal to those associated with $1$. After all, $P$’s kernel is the orthogonal complement of $V$. Since all elements of $V$ are eigenvectors of $P$ (including, not surprisingly, $u$ and $v$ themselves), you can certainly choose an orthogonal pair of them that span $V$ and thus end up with an orthogonal basis for the entire space that consists of eigenvectors of $P$.

Update: That being said, the operator $Q$ in the exercise isn’t, strictly speaking, a projection since $Q^2\ne Q$. The vectors $\Psi$ and $\Phi$ would have to be unit vectors for $Q$ to be a projection in this sense. Nevertheless, it does map the three-dimensional domain onto the two-dimensional span of $\Psi$ and $\Phi$, so you can expect it to have $0$ as an eigenvalue with the kernel of $Q$ as the associated eigenspace. By construction of $Q$, this space will consist of all vectors that are orthogonal to both $\Psi$ and $\Phi$, i.e., it is the orthogonal complement of their span. Moreover, eigenvectors associated with other eigenvalues must lie in the span of $\Psi$ and $\Phi$, and so will be orthogonal to this eigenspace as well.

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  • $\begingroup$ So, because the third eigenvector corresponds to an eigenvalue=0 it is a basis for null(P) while V=range(P)(the image) so it is not surprising. I thought that all three eigenvectors form a basis for V. So, it was a misunderstanding, i forgot that eigenvectors with eigenvalue=0 are not a basis for range(V) but for null(P). $\endgroup$ – TheQuantumMan May 2 '16 at 10:03

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