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# Minimal Polynomial of a Matrix | TIFR GS-2018 (Part B)

Try this beautiful problem from TIFR GS 2018 (Part B) based on Minimal Polynomial of a Matrix. This problem requires knowledge linear algebra.

Try this beautiful problem from TIFR GS 2018 (Part B) based on Minimal Polynomial of a Matrix. This problem requires knowledge of linear algebra.

## Minimal Polynomial of a matrix – TIFR GS- Part B (Problem 10)

The minimal polynomial of $\begin{pmatrix} 2 &1& 0&0 \\ 0 &2&0&0 \\ 0&0&2&0\\ 0&0&1&5\end{pmatrix}$ is

• $(x-2)(x-5)$
• $(x-2)^2(x-5)$
• $(x-2)^3(x-5)$
• None of these

### Key Concepts

Linear Algebra

Matrix / Vector Space

Characteristic Polynomial

But try the problem first…

Answer: $(x-2)^2(x-5)$

Source

TIFR GS -2018 (Part- B) | Problem No 10

Graduate Texts in Mathematics : Springer-Verlag

## Try with Hints

First hint

Some Definitions and Results Needed :

1. Monic Polynomial : A polynomial is said to be monic if the coefficient of the highest degree term is 1.
2. Characteristic Polynomial of a matrix : Let $A$ be a square matrix of order $n$ then the polynomial $|A-\lambda I_n|$ is called its characteristic polynomial and $|A-\lambda I_n|=0$ is called the characteristic equation. [$I_n$ is the identity matrix of order $n$]
3. Cayley – Hamilton Theorem : If $p(\lambda)$ is the characteristic polynomial of an $n\times n$ matrix $A$ over a field $F$, then the matrix $A$ satisfies the equation $p(x)=0$, i.e., $p(A)=0$. In other words, every square matrix satisfies its own characteristic equation.
4. Minimal Polynomial : The monic polynomial of lowest degree satisfied by a square matrix $A$ is called its minimal polynomial
5. Let $p(\lambda)$ and $m(\lambda)$ be the characteristic and minimal polynomials of a square matrix $A$ of order $n$ respectively. Then either both $p(\lambda)$ and $m(\lambda)$ are of degree $n$ or $m(\lambda)$ is a factor of $p(\lambda)$ .
6. Minimal polynomial is unique.

So all the ingredients you need to cook the problem are given… Can you make it delicious ?

Second Hint

To find the Characteristic equation of the given matrix :

Let $A= \begin{pmatrix} 2 &1& 0&0 \\ 0 &2&0&0 \\ 0&0&2&0\\ 0&0&1&5\end{pmatrix}$

Then, $|A-\lambda I_4|= \begin{vmatrix} 2-\lambda &1& 0&0 \\ 0 &2-\lambda &0&0 \\ 0&0&2-\lambda &0\\ 0&0&1&5-\lambda \end{vmatrix}$

$\quad = (2-\lambda)^3(5-\lambda) = p(\lambda)$ [say]

then, the characteristic equation of $A$ is $p(x)=(x-2)^3(x-5)=0$

Then By Cayley Hamilton Theorem $(A-2I_4)^2(A-5I_4)=O_{4 \times 4}$ [$O_{4 \times 4}$ is the NULL MATRIX of order $4$]

As minimal polynomial is unique then if minimal polynomial is a polynomial of degree $4$ it is same as the characteristic polynomial by Property 5 in the first hint and if minimal polynomial is less than degree $4$ then it is a factor of characteristic polynomial.

all the factors of characteristic polynomial are :

$p_1(x)=(x-2)(x-5),\quad p_2(x)=(x-2)^2(x-5),\\ \text{ and } p(x)=(x-2)^3(x-5)$

Final Step

Lets Find $p_1(A)$ i.e., $(A-2I_4)(A-5I_4)$ :

$(A-2I_4)(A-5I_4)=\begin{pmatrix} 0 &1& 0&0 \\ 0 &0&0&0 \\ 0&0&0&0\\ 0&0&1&3\end{pmatrix} \times \begin{pmatrix} -3 &1& 0&0 \\ 0 &-3&0&0 \\ 0&0&-3&0\\ 0&0&1&0\end{pmatrix}$

$\quad\quad= \begin{pmatrix} 0 &-3& 0&0 \\ 0 &0&0&0 \\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} \ne O_{4 \times 4}$

Now, $p_2(A)$ i.e., $(A-2I_4)^2(A-5I_4)$ :

$(A-2I_4)^2(A-5I)=\begin{pmatrix} 2 &1& 0&0 \\ 0 &2&0&0 \\ 0&0&2&0\\ 0&0&1&5\end{pmatrix}^2 \times \begin{pmatrix} -3 &1& 0&0 \\ 0 &-3&0&0 \\ 0&0&-3&0\\ 0&0&1&0\end{pmatrix}$

$\quad\quad= \begin{pmatrix} 0 &0& 0&0 \\ 0 &0&0&0 \\ 0&0&0&0\\ 0&0&3&9\end{pmatrix} \times \begin{pmatrix} -3 &1& 0&0 \\ 0 &-3&0&0 \\ 0&0&-3&0\\ 0&0&1&0\end{pmatrix} = O_{4 \times 4}$

Therefore the lowest degree monic polynomial satisfied by $A$ is $(x-2)^2(x-5)$.

Hence the minimal polynomial is $(x-2)^2(x-5)$

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