Take the following matrix:

\begin{pmatrix} 1&1/2 \\ 2&1 \end{pmatrix}

It has trace $2$.
It has equal values on the diagonal, thus it is scalar.
It is has determinant $0$, thus it is singular and non-invertable.
$AA^T ≠ I$, thus it is non-orthogonal.
$AA^* ≠ I$, thus it is non-unitary.
$AA ≠ I$, thus it is non-involutive.

Q1: Are the above properties correct?
Q2: Did I miss any other important properties?
Q3: What can we infer from the trace alone or in combination with other properties?
Q4: What else can we infer from the det=0 alone or in combination with other properties?

Here are my main questions though:

$AA = \mathrm{trace}(A) A$

Which "seems" to be make it idempotent but for the trace factor. Likewise, it "seems" to be a projection matrix but again, the trace factor ruins the strict definition.

Q5: Is matrix A a projection matrix?
Q6: If not, what is the best way to classify it?
Q7: Either way, what can we infer from it's structure? Q8: Can we say anything about eigenvalues

Further, if I calculate the eigenvalues by solving $\det(λI - A) =0$, I get $λ=0$ and $λ=2$.

Q9: What does the eigenvalue tell me about my matrix?
Q10: Is the second eigenvalue related to the fact that Trace=2?
Q11: What does this tell me about my eigenvectors?

  • 1
    $\begingroup$ Equal values on the diagonal does not make it a scalar matrix - a scalar matrix has equal values on the diagonal and $0$s off the diagonal. $\endgroup$ – ajd Nov 30 '14 at 20:12
  • $\begingroup$ Another noteworthy property is that the matrix is not normal $\endgroup$ – Omnomnomnom Nov 30 '14 at 20:18
  • $\begingroup$ Thanks for clarifying the Scalar property and for adding non-normal to the list. :) $\endgroup$ – Ricardo J Rademacher Nov 30 '14 at 22:45
  • $\begingroup$ Oh, and thanks for the editing of my post Adj. :) :) $\endgroup$ – Ricardo J Rademacher Nov 30 '14 at 23:46

It's twice a projection matrix. A projection matrix will have all eigenvalues either $0$ or $1$. If you divide your matrix by $2$, that's what you have.

Geometrically, what's happening is that your matrix is performing a linear projection onto a line, then doubling the length of everything on that line.

  • $\begingroup$ In light of this, is there anything that can be said about the inner product <A,A>=Trace(A*A)= 13/4? $\endgroup$ – Ricardo J Rademacher Nov 30 '14 at 23:28

If this were a projection matrix, then in some basis we could write it as

$$B = \begin{pmatrix} 0 \ \ 0 \\ 0 \ \ 1 \end{pmatrix}$$

Can we for your matrix?


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