Are you preparing for any competitive exam of Linear Algebra or want to be expert at Linear Algebra , then this post is for you.
In this post I am going to incorporate many challenging problems from various authors and will solve it .
1). Let ℂ, ℝ, and ℚ be the fields of complex, real, and rational numbers,
respectively. Determine whether each of the following is a vector
space. Find the dimension and a basis for each that is a vector space.
(a) ℂ over ℂ.
(b) ℂ over ℝ.
(c) ℝ over ℂ.
{d) ℝ over ℚ.
(e) ℚ over ℝ.
(f) ℚ over ℤ, where ℤ is the set of all integers.
(g) S = {a + b √2 + c √5 : a, b, c ∈ ℚ} over ℚ, ℝ, or ℂ.
Solution:--
(a) Yes. Dimension is 1. {1} is a basis.
(b) Yes. Dimension is 2. {1, i} is a basis.
(c) No, since i ∈ ℂ and 1 ∈ ℝ, but i ·1 = i ∉ ℝ.
(d) Yes. Dimension is infinite, since 1, π, π², π³,....... are linearly independent over ℚ.
(e) No, since √2 ∈ ℝ and 1 ∈ ℚ,
but √2 · 1 = √2 ∉ ℚ.
(f) No, since ℤ is not a field.
(g) Yes only over ℚ the dimension is 3, and { 1, √2, √5} is a basis.
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2). Consider ℝ² over ℝ. Give an example of a subset of ℝ² that is
(a) closed under addition but not under scalar multiplication;
(b) closed under scalar multiplication but not under addition.
Solution:--
(a) All vectors with initial point O and terminal points in the first quadrant.
that is
V = { (x,y)∈ℝ² : x≥0, y≥0}
Since (1,1)∈V and -1∈ℝ But (-1)(1,1)∉V
(b) All vectors with initial point O and terminal points in the first or third quadrants.
That is
V = { (x,y)∈ℝ² : x≥0, y≥0 Or x≤0 , y≤0 }
Since
(-3,-1)∈V and (2,2)∈V
But (-3,-1)+(2,2)=(-1,1)∉V
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3). If a matrix A of size 2 by 2 satisfy A = AB -BA, Then prove that A² is zero matrix.
Solution:--
First of all we shall find trace of A.
Since A= AB - BA
A= AB−BA
⇒trace(A) = trace(AB−BA)
⇒trace(A) = trace(AB)−trace(BA)
⇒trace(A) = trace(AB)−trace(AB)
⇒trace(A) = 0
Now by Cayley−Hamilton theorem, we know that A must satisfy its characteristic equation
i.e., ch(A) = x² − trace(A). x + det(A) = 0
⇒A² −trace(A) . A + det(A) = 0 ········(*)
⇒ A² + det(A) = 0
⇒ A² = − det(A) ········ (1)
Since A = AB − BA · ·······(2)
on pre and post multiplying equation (2) by A, we get
A² = A²B − ABA ⇒ − det(A)= −det(A).B−ABA ····(3)
A² = ABA − BA² ⇒ −det(A) = ABA+det(A). B ····(4)
On adding equations (3) and (4), we get
- 2 det (A) = 0
⇒ det(A) = 0
On putting the values of det(A) and trace(A) in equation (*), we get
A² = 0
Hence A² is zero matrix.
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3). Let A be a 5 by 5 skew-symmetric matrix with entries in ℝ and B be a 5 by 5 symmetric matrix whose (i,j)th entry is the binomial coefficient (ij) for 1≤i≤j≤5. Consider a 10×10 matrix given in block form by
. Then
A) det(C) = 1 or -1 B) det(C) = 0C) trace(C) = 0. D) trace(C) = 5
Solution:---
Given that matrix A is a 5 by 5 skew symmetric matrix with entries in ℝ so by definition of skew symmetric matrix we have A = 
From the above Matrix it is clear that the trace of matrix A is zero that is
Trace(A) =0
We know that the determinant of skew symmetric matrix of odd order is always zero.
Since the matrix A has odd order 5 and hence determinant of A must be zero being a skew symmetric matrix of odd order.
Det(A)= 0
Also given that
Clearly
It is clear that the trace of the matrix B is 5.
Trace(B) =5
Hence trace(C) = trace(A)+ trace(B) =0+5=5
Thus option (C) is incorrect and option (D) is correct.
Clearly the matrix C is upper triangular Matrix and hence its determinant can be obtained by multiplying the determinants of its diagonal sub-matrices A and B
det(C) = det(A)× det(B) = 0
Thus option (A) is incorrect and option (B) is correct.
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GATE/CS/2014 (LINEAR ALGEBRA)
4) Which one of the following is true?
if the trace of a matrix is positive and the determinant of the matrix is negative then at least one of its eigenvalues is negative
if the determinant of the matrix is positive then all its eigenvalues are positive
if the product of the trace and the determinant of a matrix is positive then all its eigenvalues are positive
if the trace of a matrix is positive then all its eigenvalues are positive
Solution:--
We know that the determinant of a matrix is equal to the product of its eigenvalues and hence it is clear that at least one of the eigenvalues of the matrix must be negative so that determinant of the matrix could be negative.Thus this is correct.
This is incorrect , because If A = - I, where I is a 2 by 2 identity matrix. Then it is clear that its Characteristic equation is given by x² - trace(A).x + det(A) = 0
⇒x²- (-2).x + 1 = 0
⇒ (x+1)² = 0
Observe that its eigenvalues are -1 , -1 ,but its determinant is positive.
This is incorrect because If
Observe tha eigenvalues of this matrix are 1 and -2.
Its trace =1+(-2)= -1
Its determinant = 1.(-2) = -2
Clearly the product of the trace and determinant of this matrix A is positive, But eigenvalues are not necessarily positive.
4. This is incorrect because If
Observe that eigenvalues of this matrix are -1 and 2.
clearly trace(A) = -1+2 =1 , which is positive ,but eigenvalues are not necessarily
positive.
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AB−BA)=tr(AB)−tr(BA)=tr(AB)−tr(AB)=0.
. Then
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