Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

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Question 1. Let Mn be the vector space of all n × n matrices with real entries under usual matrix addition and scalar multiplication. Find the dimensions of the fol lowing subspaces of Mn:

1. W1 = {A ∈ Mn : A is symmetric}
2. W2 = {A ∈ Mn : A is skew-symmetric}
3. W3 = {A ∈ Mn : A is upper triangular}
4. W4 = {A ∈ Mn : A is tridiagonal}

Solution. It is to be noted that the dimensions of the above subspaces can be found just by calculating the number of entries which we can choose freely to construct a matrix of the given type. For symmetric matrices, aij = aji , so that we are not free to choose the (j, i)th entry once we choose the (i, j )th entry i.e. once we choose the (i, j )th entry, the (j, i)th entry is automatically chosen. Thus, the dimension of W1 is just the number of entries on and above the diagonal of an n × n matrix. So, dim(W1) = n + (n − 1) + (n − 2) + . . . + 2 + 1 = Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

For skew-symmetric matrices, aij = −aji and so the diagonal entries are all zero. Also, as above, the selection of (i, j )th entry finalizes the (j, i)th entry too. So, the dimension of W2 is just the number of entries above the diagonal of an n × n matrix. Hence, dim(W1) = (n − 1) + (n − 2) + . . . + 2 + 1 = Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Similarly, dim (W3 ) =  Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Now, for a tridiagonal matrix, the entries in all other positions except in the main diagonal, the first diagonal above it and the first diagonal below it are always zero. So the dimension of W4 is just the number of entries in these three diagonals of an n × n tridiagonal matrix. There are two entries each in the first and the last rows and three each in all other rows. So, dim(W4 ) = 2 + 3(n − 2) + 2 = 3n − 2.


Question 2. Let V be the subspace of M2(R) consisting of matrices such that the entries of the first row add up to zero. Write down a basis for V.
Solution.
The matrices in V will be of the form  Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Hence, one possible basis is  Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

 

Question 3. Write down a necessary and sufficient condition, in terms of a, b, c and d (which are assumed to be real numbers), for the matrix ​  Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev not to have a real eigenvalue.

Proof. Clearly, this matrix won’t have any real eigenvalues if and only if its characteristic polynomial has all imaginary zeros.
Thus, (a + d)2 < 4(ad − bc).

 

Question 4. Given that the matrix  Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev has 1 as an eigenvalue, compute its trace and its determinant.
Solution. Let the other eigenvalue be x. Then α + 3 = 1 + x and 3α − 2 = x, so that α = 2. so, trace=5 and determinant=4

 

Question 5. Let A be a non diagonal 2 × 2 matrix with complex entries such that A = A−1. Write down its characteristic and minimal polynomials.
Solution.
Since A = A−1, so det(A) = ±1. But, for A to be non-diagonal matrix, det(A) = −1. Thus, A is of the form Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRevwhere ab = 1. Since A2 = I , so the characteristic polynomial of A is x2 − 1 and the minimal polynomial is also x2 − 1.

 

Question 6. For a fixed positive integer n ≥ 3, let A be the n × n matrix defined by A = I − Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev , where J is the n × n matrix with all entries equal to 1. Which of the fol lowing statements is not true?

1. Ak = A for every positive integer k.
2. Trace(A) = n − 1
3. Rank(A) + Rank(I − A) = n
4. A is invertible.

Solution. Simple matrix multiplication reveals that A2 = A and hence it can be shown that Ak = A for every positive integer k, so that (1) is true.
The simplest thing that we can check is

Trace(A) = Trace(I ) − T race Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev  = n − 1

Hence (2) is true.
Now, since  Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRevand so rank(I − A) = 1. Also, performing the elementary operations Rn → Rn + Rn − 1 + . . . + R1 and then

Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev  it is clear that Rank(A) = n − 1 and so (3) is true.

Now, it can be seen that x = (1, 1, . . . , 1)  Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev satisfies Ax = 0, so that 0 is an eigenvalue of A. Hence A is singular and so A is not invertible. So, (4) is not true.


Question 7. Let A be a 5 × 4 matrix with real entries such that Ax = 0 if and only if x = 0, where x is a 4 × 1 vector. Then find the rank of A.
Solution.
From the question it is clear that the homogeneous system of equations Ax = 0 has only trivial solution. So, Rank(A) = the number of unknowns = 4.


Question 8. Consider the following row vectors

a1 = (1, 1, 0, 1, 0, 0), a2 = (1, 1, 0, 0, 1, 0),
a3 = (1, 1, 0, 0, 0, 1) a4 = (1, 0, 1, 1, 0, 0),
a5 = (1, 0, 1, 0, 1, 0), a6 = (1, 0, 1, 0, 0, 1)

What is the dimension of the real vector space spanned by these vectors?
Solution
. Simple elementary operations will reveal that only four of these vectors are linearly independent. So, dimension of the vector space spanned by these vectors is 4.

Question 9. Let ​ Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRevi, j = 1, 2, . . . , n for some distinct real numbers b1 , b2 , . . . , bn . Then find det(A).

 

Solution. Performing the elementary operations, Ri → Ri − R1 for each i = 2, 3, . . . , n we see that det(A) = 0. Note that det(A) may not be zero if n = 2.
Question 10. Number of matrices with real entries, of the form ​ Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev such that A2 − A = 0 is

1. only one
2. only two
3. finitely many
4. infinitely many

 

Solution. Using the equation A2 = A, it can be shown that any matrix A of the form A =  Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev where bc = a − a2 , satisfies the given condition. So, there are infinitely many such matrices.

 

Question 11. Let V be a vector space of dimension n over the field Zp where p is a prime. Then how many elements are there in V ?
Solution.
Since dim(V ) = n, so let {e1 , e2 , . . . , en } be a basis of V . Then any element of V is a linear combination of the form Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev αi ei where αi ∈ Zp . So,
the number of elements in V is pn .

 

Question 12. If T : Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev be a linear transformation defined by T (1, 0, 0) = (1, 0, 1), T (0, 1, 0) = (0, 0, 1) and T (0, 0, 1) = (1, 0, 0). Then find the range space of T , null space of T , rank and nul lity of T

Solution. Clearly, T (x, y, z ) = xT (1, 0, 0)+yT (0, 1, 0)+zT (0, 0, 1) = (x+z , 0, x+ y) = (x + z )(1, 0, 0) + (x + y)(0, 0, 1). Thus the range space of T is {(x, 0, y) : x, y ∈ R}. Also, T (x, y, z ) = 0 ⇒ x = −y = −z and so the null space of T is {(x, −x, −x) : x ∈ R}. Rank(T )= Dimension of the range space=2 and nullity(T )=Dimension of the null space of T =1.


Question 13. Let A ∈ M2(R) such that tr(A) = 2 and det(A) = 3. Write down the characteristic polynomial of A−1 .

Solution. Here, the characteristic polynomial of A is x2 −(sum of eigenvalues)x+ (product of eigenvalues) i.e x2 − tr(A)x+ det(A). By Cayley Hamilton Theorem,

A2 − tr(A)A + det(A)I = 0

⇒A2 − 2A + 3I = 0

Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Thus the characteristic polynomial of   Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Question 14. Let a 3 × 3 matrix A have eigenvalues 1, 2, −1, then find the trace of the matrix B = A − A−1 + A2 .

Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

 

Question 15. The eigenvectors of a 3 × 3 matrix A corresponding to the eigenvalues 1, 1, 2 are (1, 0, −1)T , (0, 1, −1)T and (1, 1, 0)T . Find the matrix A.
Solution.
Let A = Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev Then, we can compute the unknowns using  

Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

 

Question 16. What is the null space of the matrix A = Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Solution. Its easy to see that the rank of A is 2 and so by rank-nullity theorem, the nullity of A is 0, so that the null space of A is the zero space.

Question 17. Let V be the vector of all n × n matrices over the field Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev of real numbers. A subset W of V is not a subspace of V if W consists of all matrices for which 

1. A2 = A
2. A2 ≠ A
3. A = AT
4. A = −AT

Solution. Clearly, under the restriction A2 ≠ A, the zero vector cannot belong to W and so W cannot be a subspace of V .

 

Question 18. Let A be a 4 × 4 invertible matrix. Which of the following is NOT true?

1. The rows of A form a basis of Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev
2. Null space of A contains only the zero vector
3. A has four distinct eigenvalues
4. Image of the linear transformation x → Ax on
Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev 

Solution. Since A is invertible, so rank(A) = 4 and so the rows of A are linearly independent and hence they form a basis of Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev . Since nullity(A) = 0, so the null space of A contains only the zero vector. It can be seen that for e1 = (1, 0, 0, 0)T , e2 = (0, 1, 0, 0)T , e3 = (0, 0, 1, 0)T , e4 = (0, 0, 0, 1)T , their images Ae1, Ae2, Ae3 and Ae4 are the columns of A, which are linearly independent as rank(A) = 4. But it is not necessary that A has distinct eigenvalues. Thus option 3 is NOT correct.


Question 19. Let C Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev be the linear space of all continuous functions from R to R. Let Wc be the set of differentiable functions u(x) that satisfy the differential equation u′ = xu + c. For which value(s) of the real constant c is this set a linear subspace of C (Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev)? Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Solution. Clearly if Wc is a linear subspace then u + v ∈ Wc whenever u, v ∈ Wc , so that u + v should also satisfy the given differential equation. So, u′ = xu + c, v′ = xv + c and u′ + v′ = x(u + v) + c. Hence, 2c = c ⇒ c = 0.

Question 20. Let U and V both be two-dimensional subspaces of Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev and U ≠ V , and let W = U ∩ V . Find all possible values for the dimension of W.

Solution.:  Here, dim(U ) = 2 and dim(V ) = 2. Let BU = {e1, e2} and BV = {e3, e4 } be bases of U and V respectively. It may happen that {e1 , e2 , e3, e4 } is LI, in which case dim(U + V ) = 4 ⇒ dim(U ) + dim(V ) − dim(U ∩ V ) = 4 ⇒ dim(U ∩ V ) = 0.

Again, it may happen that {e1 , e2 , e3, e4 } is LD but {e1, e2 , e3 } is LI, in which case dim(U + V ) = 3 ⇒ dim(U ∩ V ) = 1.


Question 21. Let Pn be the vector space of all polynomials of degree at most n ≥ 6 with real coefficients. Consider the following subspaces of Pn :

W1 = {p(x) ∈ Pn : p(1) = 0, p(2) = 0, p(3) = 0, p(4) = 0}

W2 = {p(x) ∈ Pn : p(3) = 0, p(4) = 0, p(5) = 0, p(6) = 0}

Find the dimensions of W1, W2 and W1 ∩ W2

Solution. Here,

W1 = {p(x) ∈ Pn : p(x) = (x − 1)(x − 2)(x − 3)(x − 4)q(x) where q(x) ∈ Pn−4 }

W1 = {p(x) ∈ Pn : p(x) = (x − 3)(x − 4)(x − 5)(x − 6)q(x) where q(x) ∈ Pn−4 }

Thus, dim(W1 ) = dim(W2) = n − 3. Now, W1 ∩ W2 contains all polynomials of the form
p(x) = (x − 1)(x − 2)(x − 3)(x − 4)(x − 5)(x − 6)q(x), where q(x) ∈ Pn−6
Thus, dim(W1 ∩ W2 ) = n − 5


Question 22. Let P3 denote the real vector space of all polynomials with real coefficients having degree less than or equal to 3, equipped with the standard ordered basis {1, x, x2 , x3 }. Write down the matrix, w.r.t this basis,of the fol lowing linear transformation:

L(p) = p′′ − 2p′ + p, p ∈ P3

Also, find the unique polynomial p such that L(p) = x3.
 

Solution. Here,

Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Thus matrix of L = Questions:Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics Physics Notes | EduRev

Now let p = ax3 + bx2 + cx + d ∈ P3 be such the L(p) = x3 . Then, using the definition of L and equating the coefficients on both sides, we get p = x3 + 6x2 + 18x + 24.

 

Question 23. A non-zero matrix A ∈ Mn(R) is said to be nilpotent if Ak = 0 for some positive integer k ≥ 2. If A is nilpotent, which of the fol lowing statements are true? 1. k ≤ n for the smal lest such k. 2. The matrix I + A is invertible. 3. Al l the eigenvalues of A are zero.


Solution : Clearly, Ak has all its eigenvalues equal to zero. But eigenvalues of Ak are just the kth power of the eigenvalues of A. Thus, all the eigenvalues of A must be zero. Hence (3) is true. The eigenvalues of I + A are 1 + 0, 1 + 0, 1 + 0 i.e. 1, 1, 1 and so it is invertible. In fact (I +A)−1 = I −A+A2 −A3 +. . .+(−1)k−1Ak .
Hence, (2) is true. Since all the eigenvalues of A are zero, so the characteristic equation of A is xn = 0 and so by Cayley Hamilton theorem, An = 0. Thus, the smallest such k for which Ak = 0 can at most be equal to n. Thus, k ≤ n.
Hence, (1) is also true.

 

Question 24. Justify whether the fol lowing are true or false: 1. There exist n × n matrices A and B such that AB − BA = I .

2. Let A and B be two arbitrary n × n matrices. If B is invertible, then tr(A) = tr−1 (B − 1AB) where tr(M ) denotes the trace of an n × n matrix M.

Solution. (1) is false. Suppose there exist such matrices. Then tr(AB − B A) = tr(I ) ⇒ tr(AB) − tr(BA) = n. But tr(AB) = tr(BA) and so we would end up with 0 = n which is absurd. Hence there cannot exist such matrices. (2) is true. This is because similar matrices have same eigenvalues and here A and B−1 AB are similar. Thus, tr(A) = sum of the eigenvalues = tr(B−1 AB)

 

Question 25. Let V be the real vector space of all polynomials in one variable with real coefficients and of degree less than, or equal to 5. Let W be the subspace defined by W = {p ∈ V |p(1) = p′(2) = 0}. What is the dimension of W ?
Solution.
Since p(1) = 0, so (x − 1) is a factor of p(x) ∈ V . thus, p(x) = (x − 1)q(x) where deg(q(x)) ≤ 4. Now, p′(2) = 0 implies that q(2) + q′(2) = 0.
Now, q(x) = (x − 2)g(x) + q(2) where deg(g(x)) ≤ 3.
Thus, the elements in W are of the form p(x) = (x − 1)(x − 2)g(x) + q(2)(x − 1).
But since W is a subspace, so we must have q(2) = 0. Thus, q′(2) = 0 and any element of W is of the form p(x) = (x − 1)(x − 2)g(x), where g(x) is a polynomial of degree at most 3. So, dimension of W is 4. 

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