§3.3
9. Consider the vectors cos(x + α) and sin(x) in C[−π, π]. For what values
of α will the two vectors be linearly dependent? Give a graphical interpretation
of your answer.
Solution:
Let us first define a matrix A as:
sin(x) cos(x + α)
A=
cos(x) −sin(x + α)
We know from theorem 3.3.3 that these vectors will be linearly independent
when det(A) 6= 0. Therefore we need to consider the values of α that will make
det(A) = 0.
We know:
det(A) = −sin(x)sin(x + α) − cos(x)cos(x + α)
= −(cos(x)cos(x + α) + sin(x)sin(x + α))
= −cos(x − (x + α))
= −cos(α)
Since −cos(α) = 0 when α is an odd multiple of π/2, we will consider those
values of α. Clearly when α is an odd multiple of π/2, cos(x) is shifted to the
left or right by an odd multiple of π/2. When this occurs cos(x + α) becomes
equivalent to sin(x) or −sin(x).
sin(x) and sin(x) are clearly linearly dependent becuase the system:
c1 sin(x) + c2 sin(x) = 0
has the nontrivial solution c1 = 1 and c2 = −1.
Similarly, sin(x) and −sin(x) are linearly dependent becuase the system
c1 sin(x) − c2 sin(x) = 0
has the nontrivial solution c1 = 1 and c2 = 1.
Thus, both vectors are clearly linearly dependent when α is an odd multiple
of π/2 (Note that the graphical interpretation is above).
13. Prove that any nonempty subset of a linearly independent set of vectors
{v1 , ..., vn } is also linearly independent.
Solution:
I will prove the contrapositive.
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If we have a nonempty subset of a set of vectors {v1 , ..., vk }, k ≤ n, and this
subset is linearly dependent, then we have that for some c1 , c2 , ..., ck ,
c1 v1 + c2 v2 + ... + ck vk = 0
where c1 , c2 , ..., ck do not all equal 0.
Now, let’s look at the set of vectors {v1 , ..., vn } from which we have the aforementioned subset. This set has at least the members contained in our subset.
Therefore, it has values of c1 , c2 , ..., ck , ..., cn , k ≤ n, other than c1 = c2 = ... =
ck = ... = cn = 0 that will make the equation
c1 v1 + c2 v2 + ... + ck vk + ... + cn vn = 0
Since we have proven the contrapositive of our original statement, we have that
our original statement is also proven, which is our desired result.
14. Let A be an m × n matrix. Show that if A has linearly independent column
vectors, then N (A) = {0}
Solution:
I will call the column vectors of A: a1 , a2 , . . . an . Since the column vectors of A are linearly independent we know:
c1 a1 + c2 a2 + . . . + cn an = 0
only has the solution c1 = c2 = . . . = cn = 0 by definition.
Recall that N (A) = {x Rn : Ax = 0}. This means that for any x in N (A) we
must have Ax = 0. Let us represent any x Rn by x = (c1 , c2 , . . . cn )T where
ci is a scalar. We already know that:
c1 a1 + c2 a2 + . . . + cn an = 0 implies
c1 = c2 = . . . = cn = 0.
Then, by the hint given in the book we know that:
Ax = c1 a1 + c2 a2 + . . . + cn an .
Thus, we know that Ax = 0 implies that x = (0, 0, . . . 0)T which clealry means
that N (A) = {0}.
15. Let {x1 , ..., xk } be linearly independent vectors in Rn , and let A be a nonsingular n × n matrix. Define yi = Axi for i = 1, ..., k. Show that {y1 , ..., yk }
are linearly independent.
Solution:
We have that {x1 , ..., xk } are linearly independent vectors, so we then know that
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the n×n matrix x1 ... xk is invertible. We also have that A is nonsingular,
so we know that A is invertible. Therefore, the product A x1 ... xk is
invertible as well. Simplifying the product,
we obtain Ax1 ... Axk which
is equal to the matrix y1 ... yk . Since we have that all yi = Axi for
i = 1, ..., k can be written as an n × n invertible matrix, we also have that the
column vectors {y1 , ..., yk } are linearly independent, our desired result.
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