Linear Algebra- Miscellaneous

Linear Algebra- Miscellaneous

M. Velrajan

 

1.

2.

Clearly T ≠ I, T2 = I. Hence T7 = T ≠ I, 3 is false and 4 is true. And (x - 1)(x + 1) is the minimal polynomial of T, hence 1, - 1 are the only eigenvalues of T and at least one with geometric multiplicity 1. The matrix of T w.r.t. the standard basis is

m(T) = 0 0 0 0 0 0 1

             0 0 0 0 0 1 0

             0 0 0 0 1 0 0

             0 0 0 1 0 0 0

             0 0 1 0 0 0 0

             0 1 0 0 0 0 0

             1 0 0 0 0 0 0 

m(T) can be obtained from the identity matrix I by interchange of 3 pairs of rows(1 ↔ 7, 2 ↔ 6, 3 ↔ 5). Hence det(T) = det(m(T)) = (-1)3det(I) = - 1. 

1 is false. 

The product of eigenvalues of T = det(T) = -1 and 

the sum of eigenvalues of T = tr(T) = 1. 

Since 1, - 1 are the only eigenvalues of T and dim(R7) = 7, 

algebraic multiplicity of 1 = 4 and algebraic multiplicity of - 1 = 3. Hence T is not diagonalizable and 2 is false.  Ans. 4.

 

3.

By properties, Ans.  3, 4. 

4.

 

By direct calculation, 2 is true.       

Tw(Tw(v)) = v for all v, hence Tw = (Tw)-1, 3 is true and det(Tw) = 1 or -1. 

If {e1, . . . , en} is the standard basis of Rn, then 

Te1(ei) = -e1, if i = 1 and ei, otherwise, hence det(Te1) = -1. So, 1 is false.  

T2w(v) = Tw(v) for all v, hence 4 is false.     Ans. 2, 3.

 

5.

The space of solutions of an non-homogeneous system of equations Ax = b (where b ≠ 0) can be always written as x0 + x, where x0 is some (arbitrary) particular solution satisfying Ax0 = b and x is a solution for Ax = 0.   The dimension of space of solutions = dim ker⁡A = dim NullspaceA = nullity A.

In this case, we have {x | Ax = b} = [1, 2, 3, 4]T + span{[2, 3, 4, 5]T}. 

Note that [1, 2, 3, 4]T is a particular solution of Ax = b. 

Hence Null space A = ker⁡A = span{[2, 3, 4, 5]T}, so nullityA = 1 and hence rankA = 4 - 1 = 3.     Ans. 2.

 

6.

 

From the minimal polynomial, size of a block corresponding to eigenvalue 2 should not be more than 2 and size of a block corresponding to eigenvalue to 3 should not be more than 1, hence 1 and 4 are not possible.            Ans. 2, 3.

7.

Recall Jordan canonical form. 

 From the given characteristic polynomial, 2 is the only char root of T. Since the deg min poly is 2,  there should be one Jordan block of size 2 corresponding to the char root 2. Since deg char. poly is 4, the possible canonical blocks sizes are 2+2 and 2+1+1. So (1) and (2) can be Jordan canonical forms of T.  (3) has no block of order 2 and in (4) the second block in the third row is not a Jordan block.                                       Ans. (1), (2).

8.

Since A is real symmetric the eigenvalues of A are real.

The eigenvalues of B = I + iA are 1 + i 𝛂, 𝛂 is an eigenvalues of A. Since 𝛂 ∈ ℝ, 1 + i 𝛂  ≠ 0 and 1 + i 𝛂 ∈ ℝ only when 𝛂 = 0. 

Hence 0 is not an eigenvalue of B and det(B)  ≠ 0. 4 is true and 1 is false. 

All the eigenvalues of a real symmetric matrix need not be all zero. 2 is false.

det(B - I) = det(iA) = det(A), so B - I is invertible iff A is invertible, 3 is false.

Ans. 4.

9.

A2 = (-1  1; -1  0) ,  A3 = (-1  0; 0  -1), A6 = I. Ans. 4.  

10.

Ans. 4. 

 11.

𝛟(v, w) = v1w1 + 2v1w2 + 4v2w1 + 3v2w2 . 1 and 2 are false.

Since 𝛟(v, w) is real bilinear, 𝛟(v, v) is a real quadratic form and 3 is true.

(or) 𝛟(v, v) = v1v1 + 6v1v2 + 3v2v2 = vTBv, B = (1  3; 3  3). 3 is true.

Since 𝛟(v, w) is bilinear,  𝛙 is not linear. 4 is false. Ans. 3.

12.

Since A is real symmetric the eigenvalues of A are real.

If A has distinct eigenvalues then A is diagonalizable over R and hence over C. 2 and 3 are true.

Eigenvalues of A = (1  0  1; 0  1 0; 0  0  2)3 x 3 are 1 and 2 with algebraic multiplicity 2 and 1 respectively, but geometric multiplicity of 1 = 1.

Hence A is not diagonalizable over C (or R). 1 and 4 are false.

Ans. 2, 3.

13.

A = 0  1  0  0

       0  0  2  0

       0  0  0  3

       0  0  0  0

0 is the only eigenvalues of A. rank(A) = 3, nullity(A) = 4 - 3 = 1.

Geometric multiplicity of 0 = 1  ≠  4 = algebraic multiplicity of 0.

1 and 4 are true and 2 and 3 are false.   Ans. 1, 4.

14.

Since A is real symmetric the eigenvalues of A are real. Since A is non singular

the eigenvalues λ1, λ2, λ3, λ4 of A are non zero real. Clearly there exists positive integer p such that  p + λ1, p + λ2, p + λ3, p + λ4 are positive and hence pI + A is positive definite (p + λ1, p + λ2, p + λ3, p + λ4 are the eigenvalues of pI + A).

λ12, λ22, λ32, λ42 are the eigenvalues of A2and hence A2 is positive definite.

1/λ12, 1/λ22, 1/λ32, 1/λ42 are the eigenvalues of A- 2and hence A - 2 is positive definite.

If A is the matrix with all the entries 1 then 0 is an eigenvalue of A, hence for all positive integer p, 1 is an eigenvalue of exp(pA) and 0 is an eigenvalue of  exp(pA) - I.  4 is false.

Ans. 1, 2, 3.

15.

AAt is an m × m symmetric matrix and rank(AAt) = rank(AtA) = rank(A) = m.

Since xtAAtx = 𝛂 xtx = 𝛂 (x12+· · ·+xm2 ), 𝛂 is the only eigenvalue of AAt of algebraic multiplicity m.

AtA is an n × n symmetric matrix and rank(AtA) = m < n. Hence det(AtA) = 0 and 0  is an eigenvalue of AtA.

Let A and B be m × n and n × m matrices and n ≥ m. Then AB and BA have the same non zero eigenvalues with the same algebraic multiplicity and if 0 is an eigenvalue of AB with algebraic multiplicity k ≥ 0 then 0 is an eigenvalue of BA with algebraic multiplicity k + n -  m.

Hence 𝛂 is the only non zero eigenvalue of AtA of algebraic multiplicity m and hence 0 is an eigenvalues of AtA of algebraic multiplicity n - m.

Ans. 1, 2, 3.

16.

Given the linear system Ax = b has a solution, for each b in Rm, hence n ≥ m and the rank of the augmented matrix (A|b) = rank(A) = r  = m. 1 is true.

dim(Column space A) =  rank(A) = r  = m. 2 is false.

If m = n then dim(null space A) = m - rank(A) = 0. 3 is false.

If m ≥ n then since n ≥ m, m = n.  4 is true.     Ans. 1, 4.

 

17.

Which of the following statements are true, for 𝛂 ∈ ℝ?

1. If 𝛂3 is algebraic over ℚ, then 𝛂 is algebraic over ℚ 

2. 𝛂 could be algebraic over ℚ[2] but may not be algebraic over ℚ

3. 𝛂 need not be algebraic over any subfield of ℝ

4. There is an 𝛂 which is not algebraic over ℚ[-1]

If 𝛂3 satisfies the polynomial a0 + a1x + a2x2 + . . .  + anxn, over ℚ, then 𝛂 satisfies the polynomial a0 + a1x3 + a2x6 + . . .  + anx3n, hence 1 is true.

Since any 𝛂 ∈ ℝ satisfies the polynomial x - 𝛂, over ℝ, 𝛂 is algebraic over ℝ. Since ℝ itself is a subfield of ℝ, 3 is false.

If 𝛂 ∈ ℝ satisfies the polynomial 

(a0 + b0-1) + (a1 + b1-1)x + (a2 + b2-1)x2 + . . .  + (an + bn-1)xn, over ℚ[-1] then (a0 + b0-1) + (a1 + b1-1)𝛂 + (a2 + b2-1)𝛂2 + . . .  + (an + bn-1)𝛂n = 0.

i.e. (a0 + a1𝛂 + a2𝛂2 + . . .  + an𝛂n) + (b0 + b1𝛂 + b2𝛂2 + . . .  + bn𝛂n)-1 = 0

Hence a0 + a1𝛂 + a2𝛂2 + . . .  + an𝛂n = 0 and b0 + b1𝛂 + b2𝛂2 + . . .  + bn𝛂n = 0.

So, 𝛂 is algebraic over ℚ. 

Hence any 𝛂 ∈ ℝ which is not algebraic over ℚ, is not algebraic over 

ℚ[-1]. 4 is true.

Ans. 1, 2, 4.

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Linear Algebra - Miscellaneous