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Hi so I have this matrix $$\left[\begin{matrix} 0 & 4 & 3\\4& 1& 0\\3& 0 & 2\end{matrix}\right]$$ that im trying to find if its digonalizable or not but i cant seem to find the proper eigen values because one of it comes out to be imaginary. What am I doing wrong?

antara roy
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    Hint: Notice that the matrix is symmetric. What does that tell you about its diagonalizability? – Prasun Biswas Oct 27 '17 at 13:12
  • Eigenvalues can be imaginary, but when the entries in the matrix are integers, imaginary or complex roots come in pairs. – Mark Bennet Oct 27 '17 at 13:15
  • @MarkBennet Doesn't this have more to do with the fact that the entries are real? I mean you are technically right, but it sounds strange to reason that it is because they are integers. – M. Winter Oct 27 '17 at 13:21
  • @M.Winter Indeed real entries is also correct. I learned about matrices before I learned in any rigour about real numbers (strange as it may seem). So I wasn't sure what background I could assume. – Mark Bennet Oct 27 '17 at 13:25
  • Yes it should be real entries. The reason is, that then the coefficients of the characteristic polynomial are real. So by complex-conjugation of the polynomial you can deduce that if $z$ is a root, also $\overline{z}$ is a root. – P. Siehr Oct 27 '17 at 13:25
  • It seems that all the eigenvalues are real: https://www.wolframalpha.com/input/?i=caracteristic+equation{{0,4,3},{4,1,0},{3,0,2}} – Emilio Novati Oct 27 '17 at 13:25
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    @EmilioNovati They have to be real, see the answer below. – P. Siehr Oct 27 '17 at 13:26
  • Yes, I agree that ''they have not to be'', but they are :) – Emilio Novati Oct 27 '17 at 13:27
  • Hint taken, added and have real eigenvalues to the answer. Thanks^^ – P. Siehr Oct 27 '17 at 13:30
  • I'm still unsure where I'm going wrong. What do the eigen values come out to be? – antara roy Oct 27 '17 at 13:33
  • So why don't you tell us step-by-step, what you did by editing your question? – P. Siehr Oct 27 '17 at 13:38

3 Answers3

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Since you did not tell us, where you are stuck, we can't help you with what you struggle. Hence this answer is minimalistic, but complete.

Real, symmetric matrices are diagonalizable (by orthogonal matrices) and have real eigenvalues. And your matrix is real and symmetric.

Have a look at the spectral theorem.

P. Siehr
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  • in case of interest: sometimes it becomes clear that the student is being asked about congruence rather than similarity (and may post contradictory phrases), see example http://math.stackexchange.com/questions/395634/given-a-4-times-4-symmetric-matrix-is-there-an-efficient-way-to-find-its-eige – Will Jagy Oct 27 '17 at 18:22
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    In this case the question is probably homework or from an (old) exam. It is a typical example, where the calculation of eigenvalues gets pretty ugly, as you can see in the other solution, but the task can be solved in one line. – P. Siehr Oct 27 '17 at 19:51
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I wonder if you're doing the eigenvalue calculation correctly. It should look something like this:

$$\begin{align} \det\left[\begin{matrix}-\lambda & 4 & 3 \\ 4 & 1- \lambda & 0 \\ 3 & 0 & 2-\lambda \end{matrix}\right] &= 3\left|\begin{matrix}4 & 3 \\ 1-\lambda & 0\end{matrix}\right| + (2-\lambda)\left|\begin{matrix}-\lambda & 4 \\ 4 & 1-\lambda\end{matrix}\right|\\ &=3(3(\lambda-1)) +(2-\lambda)((\lambda^2-\lambda)-16)\\ &=9\lambda-9-\lambda^3+3\lambda^2+14\lambda-32 \\ &=-(\lambda^3-3\lambda^2-23\lambda+41) \end{align}$$

In the first step there, I expanded along the bottom row, because there was a $0$ there. Your steps getting there may differ, but you should arrive at the same polynomial, which has 3 real roots, although they are not nice, rational numbers. They appear to equal approximately $-4.38, 1.62$ and $5.76$.

Double-check your calculation and see if you get the same eigenvalues.

G Tony Jacobs
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  • in case of interest: sometimes it becomes clear that the student is being asked about congruence rather than similarity (and may post contradictory phrases), see example http://math.stackexchange.com/questions/395634/given-a-4-times-4-symmetric-matrix-is-there-an-efficient-way-to-find-its-eige – Will Jagy Oct 27 '17 at 18:22
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For a symmetric matrix $H,$ without bothering about the eigenvalues, the matrix is congruent to a diagonal matrix, meaning there is a matrix $P$ with $\det P \neq 0$ such that $P^T HP = D$ is diagonal

$$ P^T H P = D $$ $$ Q^T D Q = H $$ $$ H = \left( \begin{array}{rrr} 0 & 4 & 3 \\ 4 & 1 & 0 \\ 3 & 0 & 2 \\ \end{array} \right) $$

==============================================

$$\left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 4 & 0 \\ 4 & 0 & 3 \\ 0 & 3 & 2 \\ \end{array} \right) $$

==============================================

$$\left( \begin{array}{rrr} 1 & - 4 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & - 4 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 4 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 16 & 3 \\ 0 & 3 & 2 \\ \end{array} \right) $$

==============================================

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 3 }{ 16 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 0 & 1 & \frac{ 3 }{ 16 } \\ 1 & - 4 & - \frac{ 3 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 4 & 1 & 0 \\ 1 & 0 & - \frac{ 3 }{ 16 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & \frac{ 41 }{ 16 } \\ \end{array} \right) $$

==============================================

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & - 4 & 0 \\ \frac{ 3 }{ 16 } & - \frac{ 3 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 4 & 3 \\ 4 & 1 & 0 \\ 3 & 0 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 1 & \frac{ 3 }{ 16 } \\ 1 & - 4 & - \frac{ 3 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & \frac{ 41 }{ 16 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 4 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & - \frac{ 3 }{ 16 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & \frac{ 41 }{ 16 } \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 1 & 0 \\ 1 & 0 & - \frac{ 3 }{ 16 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 4 & 3 \\ 4 & 1 & 0 \\ 3 & 0 & 2 \\ \end{array} \right) $$

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As happens sometimes, there is a more attractive version available, all integers:

$$ P^T HP = D $$

$$\left( \begin{array}{rrr} 1 & -1 & 1 \\ -2 & 1 & -1 \\ 11 & -3 & 4\\ \end{array} \right) \left( \begin{array}{rrr} 0 & 4 & 3 \\ 4 & 1 & 0 \\ 3 & 0 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & -2 & 11 \\ -1 & 1 & -3 \\ 1 & -1 & 4 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 41 \\ \end{array} \right) $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} -1 & -1 & 0 \\ 3 & 7 & 1 \\ 5 & 8 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 41 \\ \end{array} \right) \left( \begin{array}{rrr} -1 & 3 & 5 \\ -1 & 7 & 8 \\ 0 & 1 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 4 & 3 \\ 4 & 1 & 0 \\ 3 & 0 & 2 \\ \end{array} \right) $$

Will Jagy
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