Ytukyg

$$ A = begin{bmatrix} a & 0 & 0 \ b & 0 & 0 \ 1 & 2 & 1 end{bmatrix} $$ for which values of a , b the matrix is diagonizable? in each case that $ A $ is diagonizable, write similar diagonal matrix.

My attempt:

I split the solution into six cases.

The characteristic polynomial of such matrix will be $ p(x) = (a - lambda)(1- lambda) lambda $ and therefore the eigen values should be $ a , 1, 0 $.

1) If $ a = b $ and $ a not = 0,1 Rightarrow $ 3 eigen values: $lambda = 0,1,a $

$ lambda = a, |I_a - A| = begin{vmatrix} 0 & 0 & 0 \ -a & a & 0 \ -1 & -2 & a-1 end{vmatrix} Rightarrow$ matrix rank is $ 2 $ and therefore one eigen vector.

$ lambda = 1 , |I - A | = begin{vmatrix} -a & 0 & 0 \ -a & 1 & 0 \ -1 & - 2 & 0end{vmatrix} Rightarrow $ matrix rank is 2 again and therefore only one eigen vector

$ lambda = 0, |0-A| = begin{vmatrix} -a & 0 & 0 \ -a & 0 & 0 \ -1 & -2 & - 1 end{vmatrix} Rightarrow $ matrix rank is 2 again and therefore only one eigen vector

Therefore if $ a = b $ and $ a not = 0 ,1 $ then the matrix is diagonizable and similar to the matrix $ begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & a end{bmatrix} $

2) $ a = b $ and $ a = 0 $ then $ |0-A| = begin{vmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ -1 & -2 & -1 end{vmatrix} Rightarrow $ Matrix rank is one and therefore eigen value$ 0$ has two eigen vectors and the eigen value $ 1 $ must have at least one eigen vector because geometric multiplicity is always equal or greater then algebraic multiplicity. Therefore the matrix is diagonizable and similar to $ begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 end{bmatrix} $

3) If $ a= b $ and $ a= 1 $ then $ |I - A | = begin{vmatrix} 0 & 0 & 0 \ -1 & 1 & 0 \ -1 & -2 & 0 end{vmatrix} Rightarrow $ algebraic multiplicity is $ 2 $ while geometric is only one and therefore the matrix is not diagnosable.

4) $ a not = b $ and $ a not = 0,1 $ then $ A $ has three different eigenvalues and each one must have one eigenvector and because geometric multiplicity must be at least one and cannot exceed algebraic multiplicity and there the matrix is diagonizable and similar to $ begin{bmatrix} a & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 end{bmatrix} $

5) $ a not = b $ and $ a = 0 $ then $ 0 $ is eigen value with algebraic multiplicity of $ 2 $ and so $ |0 - A| = begin{vmatrix} 0 &0 & 0 \ b & 0 & 0 \ -1 & -2 & -1 end{vmatrix} Rightarrow$ matrix rank is two and therefore algebraic multiplicity > geometric multiplicity and the matrix cannot be diagonalised.

6) $ a not = b $ and $ a = 1 $ then

$ |I-A| = begin{vmatrix} 0 & 0 & 0 \ b & 1 & 0 \ -1 & -2 & 0 end{vmatrix} Rightarrow $ matrix will be diagonizable only if $ b = 1/2 $ and then geometric multiplicity of eigen value $ 1 $ will be $ 2 $ and then it will be similar to $ begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 end{bmatrix} $

Thank you, for those of you who made it so far. It is a question I had on a test and I got for this answer 2/14 points. After talking to friends who also studied the subject they told me that even though the answer may be unnecessarily long, they couldn't find any flaws.

Overall it looks correct.

There is an error (almost certainly unintentional) in part 2, where you switch geometric and algebraic multiplicity when you say "geometric multiplicity is always equal or greater then algebraic multiplicity."

Also you wrote "diagnosable" once (in 3) and "diagonizable" (which I didn't actually catch, user376343 did in the comments) all the time instead of "diagonalizable."

The rest looks mathematically correct though.

That said. The proof is stylistically wrong in many other places.

The argument in (4) applies to case (1) as well, and should've been merged.

Also you should've simplified and broken the argument into at most three cases. I'll give one suggestion for how to write this proof below.

My version

Since the matrix is lower triangular, its eigenvalues are $a,0$, and $1$. Hence if it is diagonalizable it will be similar to
$$begin{pmatrix} a & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 end{pmatrix},$$ since similar matrices have the same eigenvalues.

If $ane 0,1$ then the matrix has three distinct eigenvalues, and is thus diagonalizable.

Otherwise if $a=0$, in order to be diagonalizable, $A$ must have two $0$-eigenvectors, which means its kernel must have size 2, so it must be rank $1$. when $a=0$, $A$ has rank 1 if and only if $b=0$.

Similarly if $a=1$, in order to be diagonalizable, $A-I$ must have rank $1$.
$$A-I = begin{pmatrix} 0 & 0 & 0 \ b & -1 & 0 \ 1 & 2 & 0 end{pmatrix},$$
which has rank 1 if and only if the last two rows are linearly dependent, which is true if and only if the lower left $2times 2$ minor vanishes. I.e. if $a=1$, $A$ is diagonalizable if and only if $2b=1$, or $b=frac{1}{2}$.

Thus $A$ is diagonalizable precisely when one of (i) $ane 0,1$, (ii) $a=b=0$, or (iii) $a=1,b=1/2$ is true, and if diagonalizable it is similar to
$$begin{pmatrix} a & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 end{pmatrix}.$$