GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 22

Answer 1

Option by option analysis: 

• A. The rank of $B$

$B$ is a $3$-by-$3$ matrix and it has eigenvalues $0, 1$ and $2$. 

Since $B$ has a eigenvalue $0$, it is singular (non invertible). As it is a three-by-three matrix; we can say that its rank will be at most $2$. It has also two distinct nonzero eigenvalues, hence its rank is exactly $2$.

Here, we can also think like this, 

$B$ matrix has one eigenvalue $\lambda_{1} = 0$; So, it has at least $one$ free variable.

$B$ is a $3 \times 3$ matrix; so, the $rank$ of $B = 3 – 1 = 2$ 

• A. The determinant of $B^{T}B$

Since, $B$ is singular, $|B| = 0$. We can also get determinant of $B^{T}B$:

                $|B^{T}B| = |B^{T}||B| = 0$

100. The eigenvalues of $B^{T}B$ 

NOT enough information to conclude that we can find eigenvalues of $B^{T}B$ by the given eigenvalues.

500. The eigenvalues of $(B^{2} + I)^{-1}$ 

 If $p(t)$ is polynomial and $x$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $p(t)x = p(\lambda)x$.

If $\lambda$ is eigenvalue of $A$ then $\frac{1}{\lambda}$ is eigenvalue of $A^{-1}$.

So, given eigenvalues of $B$ is $0, 1$ and $2$. Hence, eigenvalues of $(B^{2} + I)^{-1}$ – 

$\frac{1}{0^{2} + 1} = 1$,  $\frac{1}{1^{2} + 1} = \frac{1}{2}$ and $\frac{1}{2^{2} + 1} = \frac{1}{5}$

$Ans: A;B;D$

Some Standard Resources:

$MIT$ $Answer$ $PDF$: https://ocw.mit.edu/courses/18-06sc-linear-algebra-fall-2011/a5b090d92c330788febc5b2092906af9_MIT18_06SCF11_Ses2.8sol.pdf 

$\text{StackExchange}$ $Logic$: https://math.stackexchange.com/questions/1758658/what-is-all-of-the-information-eigenvalues-tell-us 

$Relation$ $between$ $Rank$ $and$ $distinct$ $eigen$ $values$:  https://math.stackexchange.com/questions/2340541/relation-between-rank-and-number-of-distinct-eigenvalues 

Comments

bro 1 doubt,regarding the reason of rank(B)=2

U wrote,

as  It has also two distinct nonzero eigenvalues, hence its rank is exactly 2.

but as sir mentioned in lecs,the only relation between rank and eigenvalue is through ƛ=0
So rank of B  is 2 because wrt ƛ=0,only 1 LI eigen vector i.e 1 free variable in soln of Bx=0, hence 2 pivot cols,so rank(B)=2.

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@~^~Dbnjn Yes correct !! The $logic$ are given in ans now.

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Can you Tell me why we cant find eigen values here ,because

B$^{T}$B=B$^{2}$

Bx= $\lambda$ x

So,

B$^{2}$x=$\lambda ^{2}$ x

B$^{T}$B is basically $\lambda ^{2}$, right?

Answer 2

Option A : Rank of B can be find out with the help of the Eigen value 0 . ( Statement is true )

Option B : Det(B) = Product of eigen values = 0 * 1 * 2 = 0 . And  the determinant for BB' is also zero bcz |B| = 0 . ( Statement is true )

Option C : Eigens value of B'B can't be find with the eigen values .

Option D : ( B^2 + I) ^(-1) → You can simply put eigen values in place of B and 1 in place of I to get the new eigen values . ( Statement is True )

Option A , B ,D are right choices .

Answer 3

Sachin Mittal 1 sir here how we can know the exact rank because we don’t have characteristic equation . Also we just know that rank < 3 that’s it then how option A true

Comments

@Omi Kakadiya

watch live session -5, sir has explained how we can find the exact rank.

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@Omi Kakadiya

from that we can conclude that (p-0)(p-1)(p-2)

here power of the p is 1 that means we can have atmost one free variable so if we assume it has 0 free variable then it is 3 and if it is having 1 free variable rank is 2