WA 3: Solutions
1. Which of the following subsets of the vector space C[−1,1] of continuous
functions on the segment [−1,1] are subspaces (justify your answers)?
(a) The set Xof functions fwhich satisfy f(−1) + f(1) = 1.
(b) The set Yof functions fwhich satisfy
f(−t) = f(t) for every t ∈[−1,1].
(Such functions are called even.)
Solution. (a) The zero function f(t)≡0 satisfies f(−1) + f(1) = 0 + 0 = 0 6= 1,
thus this function doesn’t belong to the set X. Therefore, Xis not a subspace of
the vector space C[−1,1].
(b) The zero function f(t)≡0 satisfies f(−t) = f(t) = 0 for every t∈[−1,1],
thus this function belongs to the set Y.
For any two functions fand gfrom Y,
(f+g)(−t) = f(−t) + g(−t) = f(t) + g(t) = (f+g)(t) for every t∈[−1,1].
Thus f+g∈Y.
For an arbitrary f∈Yand c∈Rone has
(cf)(−t) = cf(−t) = cf(t) = (cf)(t) for every t∈[−1,1].
Thus cf ∈Y.
Finally, we conclude that Yis a subspace of the vector space C[−1,1].
2. Let Tbe a linear transformation mapping a vector space Vonto a vector
space W. Prove that the following statements are equivalent:
(a) Tis left invertible, i.e., there exists a linear transformation Swhich satisfies
S(T(x)) = xfor every x ∈V;
(b) ker T={0};
(c) Tis a one-to-one transformation.
Solution. (a)⇒(b). If ker T6={0}then there exists a vector x6= 0 such that
T(x) = 0, and we have
x=S(T(x)) = S(0) = 0,
which contradicts to the assumption that x6= 0. Therefore, ker T={0}.
(b)⇒(c). If for a vector y∈Wthere exist two vectors xand x0in Vsuch that
T(x) = T(x0) = ythen
T(x−x0) = T(x)−T(x0) = y−y=0,
i.e., x−x0∈ker T. Since ker T={0}by the assumption, we get x−x0=0, i.e.,
x=x0. This means that for an arbitrary vector y∈Wthe equation T(x) = yhas
at most one solution, i.e., Tis a one-to-one transformation.
(c)⇒(a). Since Tis one-to-one and onto, for an arbitrary y∈Wthe equation
T(x) = yhas exactly one solution x∈V. Denote this unique solution by S(y).
So, the transformation S:W→Vis well defined, and S(T(x)) = x. We only have
to show now that this transformation Sis linear.
For arbitrary vectors yand y0from W, we have S(y) = xif and only if T(x) = y,
and S(y0) = x0if and only if T(x0) = y0. Since the transformation Tis linear, we
have
T(x+x0) = T(x) + T(x0) = y+y0.
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