GO Classes 2024 | Weekly Quiz 7 | Linear Algebra | Question: 5

Answer 1

Here there are 3 distinct eigen values = 0,2,3 [to be exact : 0,2,3,3]. Hence option (D) can be determined.

Here power of eigen value, 0 is 1 hence the number of L.I eigen vectors = 1 (AM = GM) 

So, nullity = 1 and rank = (1+1+2) – 1 = 3.

Hence option (A) can be determined.

 

However, need some more clarification on options b and d!

Comments

A is clearly not diagonalizable. Assume A is diagonalizable, then A should have 4 independent eigen vectors ie Geometric multiplicity (G.M.) should  total four but

we know Geometric Multiplicity for each roots

$$\begin{array}{c|c|c|c} \lambda&\text{A.M}&\text{G.M}_{min}&{G.M}_{max} \\ \hline 0&1&1&1 \\ 2&1&1&1 \\ 3&2&1&2\\ \hline  \Sigma&4&3&4 \end{array}$$

So possible GM are 3,4 but we know A is singular, therefore 4 independent eigen vectors cannot exist.

 

So Diagonalizability can be determined for A,

Is my reasoning correct?

Edit: I was wrong.  A can be not invertible but can still have 4 independent eigen vectors eg. Projection

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How did you get this, ‘A is singular, therefore 4 independent eigen vectors cannot exist’ ?

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Check my edit. I was wrong.  Invertible property has no relation with Diagonalizability.

Answer 2

 

Here there are 3 distinct eigen values = 0,2,3 [to be exact : 0,2,3,3]. Hence option (D) can be determined.

Here power of eigen value, 0 is 1 hence the number of L.I eigen vectors = 1 (AM = GM) 

So, nullity = 1 and rank = (1+1+2) – 1 = 3.

Hence option (A) can be determined.

(copied from @chokostar’s answer)

A can be symmetric or not-symmetric which can be illustrated through following example:

$\begin{pmatrix} 0 & & \\ & 2 & \\ & & 3 \\ & & & 3 \end{pmatrix}$ => Symmetric

$\begin{pmatrix} 0 & & \\ & 2 &7&9\\ & & 3 &8\\ & & & 3 \end{pmatrix}$ => Not symmetric

* Blank entries are 0s

Hence option (B) CANNOT be determined.

Now, for A to be diagonalizable A should have 4 independent eigen vectors ie Geometric multiplicity (G.M.)

should  total four.

But here corresponding to λ = 3 there can be 1 or 2 LI vectors.

So, in total we can have either 3 or 4 LI vectors.

Thus, we cannot guarantee that A is diagonalizable.

Hence option (C) CANNOT be determined.