GO Classes DPP | GATE CS, DA | Linear Algebra | Diagonalization

Answer 1

Given $A^2-A=0$,

we get $A(A-I)=0$.

So $A$ satisfies the polynomial $x(x-1)=0$.

Hence the minimal polynomial of $A$ divides $x(x-1)$.

Now $x(x-1)$ has two distinct roots, namely $0$ and $1$. Therefore the minimal polynomial has no repeated root, and hence $A$ is diagonalizable.

So the correct answer is (D).

Also, the other options are false:

(A) is false, since $A=I$ also satisfies $A^2-A=0$.

(B) is false, since $A=0$ also satisfies $A^2-A=0$.

(C) is false, since rank$(A)$ need not be only $0$ or $1$. For example, $A=\operatorname{diag}(1,1,0,\dots,0)$ satisfies $A^2=A$ and has rank $2$. Another example, $A$ can be the identity, which has rank $9$ in this case.

Answer 2

 

Option D is true.

Answer 3

Answer: D

Answer 4

I don't know anything about these minimal polynomial stuff, i am going in a general way.

The information they have given, $A \in \mathbb{R}^{n \times n}$, such that $\mathrm{A}^2 = \mathrm{A}$.

Note: Suppose $\set{v_1, \ldots, v_n}$ be a columns of $A$. 

$\mathrm{A}\left[\begin{matrix}v_1 & v_2  & \ldots & v_n \end{matrix}\right] = \left[\begin{matrix}v_1 & v_2  & \ldots & v_n \end{matrix}\right]$

Which means the column vectors of $A$ are $\textbf{EigenVectors}$.

 

That's so Good, we got $\textbf{ n EigenVectors}$. Can i just call them Linearly Independent?  $\textbf{NO}$, we can't call the set of columns of $A$ as linearly independent.

Since $A^2 = A \implies |A|^2 = |A| \implies |A| = 0 \text{ or } |A| = 1$, when $|A| = 0$, then columns of A are no longer independent.

We only got  $\textbf{ n EigenVectors}$. But if i can  $\textbf{"n Linearly Independent EigenVectors"}$ then Yes, $A$ is Diagonalizable.

 

Let's try something, we going to take a subset $S$ of columns of $A$ such that every col of $A$ is a unique linear combination of vectors in $S$. See carefully, i am only taking only a subset of columns of A, which means the vectors i am picking are also $\textbf{EigenVectors}$ and the set $\textbf{S}$ is linearly independent. Don't forget $\textbf{S}$ is a basis of columns space of $A$.

What  is $\textbf{S}$?

1. $\textbf{S}$ is a subset of columns of $A$
2. $\textbf{S}$ is a basis of column space of $A$.
3. Vectors in $\textbf{S}$ are eigen vectors of $A$.
4. This is a key fact, Eigen Value of all these vectors in S is $\textbf{1}$.

Let $\dim \text{Col(A)} = k$, then it means $|S | = k $. Lets just give some name for vectors in $S$, $\textbf{S} = \set{u_1, \ldots, u_k}$.

 

From rank nullity theorem, $\dim \text{null(A)} = n -k$,  let $\set{t_1, \ldots, t_{n-k}}$ be a basis of $\text{null(A)}$. Then $At_i = 0_v$, for every $t_i$ in the basis of null space of $A$.

• But that's not all, every vector in the basis of null space of $A$ is an eigen vector of $A$, because $At_i = 0_v = 0\times t_i$, they all are eigen vectors of $A$, associated with eigen value $\textbf{0}$

 

$\textbf{KEY FACT:-}$ Union of linearly independent sets from different EigenSpaces is linearly independent. (This can be proved)

 

Since,

$\textbf{S} = \set{u_1, \ldots, u_k}$ is a linearly independent set from $1-\text{EigenSpace}$ and  $\set{t_1, \ldots, t_{n-k}}$ is linearly independent set from $0-\text{EigenSpace}$

 

$$\textbf{Then}$$

 

$\set{u_1, \ldots, u_k, t_1, \ldots, t_{n-k}}$ is a set of $\textbf{ n Linearly Independent Eigen Vectors of A}$.

 

We got it, so $\textbf{A is Diagonalizable}$.

 

$\text{Fact}: \textbf{ If } A^2 = A \implies \textbf{A is diagonalizable}$.

 

 

For option A, B, C

$$D = \left[\begin{matrix}1 &  0 &0 \\ 0 &1& 0 \\ 0 &0 & 0\end{matrix}\right]$$

 

This works as Counter Example. In case you wonder, this Projection matrix onto Span of first two columns of $D$.

 

 

D is the answer.