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Chapter 4: Problem 121

\text { Suppose } V=U \oplus W . \text { Show that } \operatorname{dim} V=\operatorname{dim} U+\operatorname{dim} W.

Short Answer

Expert verified
Given \(V = U \oplus W\), let \(\{u_1, u_2, \dots, u_m\}\) be a basis for \(U\) and \(\{w_1,w_2,\dots,w_n\}\) be a basis for \(W\). We create a set \(\{v_1, v_2, \dots, v_{m+n}\}\) by combining the basis vectors for \(U\) and \(W\). To show this set is linearly independent in \(V\), consider the linear combination: \((a_1 u_1 + a_2 u_2 + \cdots + a_m u_m) + (a_{m+1} w_1 + a_{m+2} w_2 + \cdots + a_{m+n} w_n) = 0\). The uniqueness of the direct sum implies that the coefficients must equal zero. Thus, \(\{v_1, v_2, \dots, v_{m+n}\}\) is a linearly independent basis for \(V = U \oplus W\), and \(\dim(V) = \dim(U) + \dim(W)\).

Step by step solution

01

Direct sum definition

Recall that if \(V = U \oplus W\), then every vector \(v \in V\) can be uniquely written as the sum of a vector from \(U\) and a vector from \(W\). In other words, \(v = u + w\) for some \(u \in U\) and \(w \in W\), and this representation is unique.
02

Bases of \(U\) and \(W\)

Let \(\{u_1, u_2, \dots, u_m\}\) be a basis for the vector space \(U\) and \(\{w_1,w_2,\dots,w_n\}\) be a basis for the vector space \(W\). Since these are bases for their respective vector spaces, they are linearly independent in those spaces.
03

Create a basis for the direct sum \(V = U \oplus W\)

We will now form a set \(\{v_1, v_2, \dots, v_{m+n}\}\) by combining the basis vectors for \(U\) and \(W\), as follows: \(v_1 = u_1, v_2 = u_2, \dots, v_m = u_m, v_{m+1}=w_1, v_{m+2} = w_2, \dots, v_{m+n} = w_n\)
04

Proving linear independence

Now, we need to show that the set \(\{v_1, v_2, \dots, v_{m+n}\}\) is linearly independent in \(V\). To do this, let's consider the following linear combination: \(a_1 v_1 + a_2 v_2 + \cdots + a_m v_m + a_{m+1} v_{m+1} + a_{m+2} v_{m+2} + \cdots + a_{m+n} v_{m+n} = 0\) which can be rewritten as: \((a_1 u_1 + a_2 u_2 + \cdots + a_m u_m) + (a_{m+1} w_1 + a_{m+2} w_2 + \cdots + a_{m+n} w_n) = 0\) Observe that since \(U\) and \(W\) are subspaces of \(V\), their sum must be in \(U \oplus W\). Therefore, the only way this equation can be true is if the linear combination of the \(u_i\)s is equal to the opposite of the linear combination of the \(w_i\)s: \(a_1 u_1 + a_2 u_2 + \cdots + a_m u_m = - (a_{m+1} w_1 + a_{m+2} w_2 + \cdots + a_{m+n} w_n)\) Due to the uniqueness of the direct sum, the only way the two linear combinations can be equal is if both have all the coefficients equal to zero. Therefore: \(a_1 = a_2 = \cdots = a_m = a_{m+1} = a_{m+2} = \cdots = a_{m+n} = 0\) This means that the set \(\{v_1, v_2, \dots, v_{m+n}\}\) is linearly independent in \(V\).
05

Constructing the direct sum with the new basis

Since the set \(\{v_1, v_2, \dots, v_{m+n}\}\) is linearly independent and spans both \(U\) and \(W\), it forms a basis for the direct sum \(V = U \oplus W\). Consequently, the dimension of \(V\) is given by the number of basis vectors, which is \(m+n\).
06

Concluding the proof

We have shown that if \(V = U \oplus W\), then there exists a basis for \(V\) consisting of \(m+n\) linearly independent vectors where \(m\) is the dimension of \(U\) and \(n\) is the dimension of \(W\). Therefore, we can conclude that \(\dim(V) = \dim(U) + \dim(W)\), proving the statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direct Sum of Vector Spaces
Understanding the direct sum of vector spaces is essential for grasping more complex concepts in linear algebra. In simple terms, the direct sum allows us to combine two or more vector spaces in a way that preserves their individual structures.

When we say that a vector space V is the direct sum of U and W, denoted by
\( V = U \oplus W \), it means that each element
\( v \in V \) can be uniquely represented as a sum of elements from U and W. Essentially, we are combining the two spaces without any overlap, ensuring that each vector in V is created by adding one vector from U and one vector from W, with no ambiguity in the process.

For example, if we have two subspaces of a vector space, where one is a plane through the origin and the other is a line through the origin that does not lie in the plane, their direct sum gives us the entire space. The ability to see VB as a composed space deepens our understanding of its structure and can be extremely valuable in applications such as solving a system of equations.
Basis of Vector Space
A basis of a vector space plays a role analogous to a skeleton in the body鈥攊t provides a structure upon which the entire space is constructed. A basis is fundamentally a set of vectors that are both linearly independent and span the vector space.

A vector space may have an infinite number of different bases, but each basis has a common property: the number of vectors, or the dimension, of the space. For a subspace U with a basis
\( \{ u_1, u_2, \dots, u_m \} \), every vector in U can be expressed as a combination of these basis vectors.

The choice of a good basis can simplify problems and calculations within a given vector space. Moreover, when we create a basis for a vector space that is a direct sum of two subspaces, like U and W in our example, we can simply unite their individual bases to form the basis for the entire space鈥攁nother beautiful intersection of structure and simplicity in linear algebra.
Linear Independence
Linear independence is a concept that strikes at the heart of what makes vectors useful and interesting. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others. In other words, each vector adds a new dimension to the space, bringing a unique quality that the others cannot provide.

To test whether a set of vectors is linearly independent, we can set up a linear combination of the vectors equal to zero:

\( a_1 v_1 + a_2 v_2 + \cdots + a_n v_n = 0 \).

If the only solution to this equation is that all coefficients
\( a_i \) are zero, the set is linearly independent. In our earlier example, showing the linear independence of a combined basis is crucial in verifying that we indeed have a valid basis, which impacts the dimensions and the overall structure of our direct sum space, V.
Dimension Theorem for Vector Spaces
The dimension theorem for vector spaces is like a guiding star that helps us navigate through the Universe of vector spaces. It provides a straightforward but powerful insight: the dimension of a vector space is the number of vectors in any basis of that space.

This theorem assures us that all bases of a particular space, no matter how different they may look, have the same number of elements. If we take our vector space V which is a direct sum of U and W, and establish a basis for V by combining the bases from U and W, the dimension theorem tells us that the dimension of V will be the sum of the dimensions of U and W.

In practical terms, this helps us solve problems and understand the structure of vector spaces, as knowing the dimension can inform us about possible solutions to equations and the behavior of linear transformations. Utilizing this theorem, we can confidently conclude the dimensions of more complex spaces built from simpler ones, which is precisely what we did in the exercise at hand.

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Most popular questions from this chapter

Find a basis and dimension of the subspace \(W\) of \(\mathbf{R}^{3}\) where (a) \(W=\\{(a, b, c): a+b+c=0\\}\), (b) \(W=\\{(a, b, c):(a=b=c)\\}\) (a) Note that \(W \neq \mathbf{R}^{3}\), because, for example, \((1,2,3) \notin W\). Thus, \(\operatorname{dim} W<3\). Note that \(u_{1}=(1,0,-1)\) and \(u_{2}=(0,1,-1)\) are two independent vectors in \(W\). Thus, \(\operatorname{dim} W=2\), and so \(u_{1}\) and \(u_{2}\) form a basis of \(W\) (b) The vector \(u=(1,1,1) \in W\). Any vector \(w \in W\) has the form \(w=(k, k, k)\). Hence, \(w=k u\). Thus, \(u\) spans \(W\) and \(\operatorname{dim} W=1\).

Prove Lemma 4.13: Suppose \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) spans \(V\), and suppose \(\left\\{w_{1}, w_{2}, \ldots, w_{m}\right\\}\) is linearly independent. Then \(m \leq n\), and \(V\) is spanned by a set of the form $$ \left\\{w_{1}, w_{2}, \ldots, w_{m}, v_{i_{1}}, v_{i_{2}}, \ldots, v_{i_{n-w}}\right\\} $$ Thus, any \(n+1\) or more vectors in \(V\) are linearly dependent. It suffices to prove the lemma in the case that the \(v_{i}\) are all not \(0 .\) (Prove!) Because \(\left\\{v_{i}\right\\}\) spans \(V\), we have by Problem \(4.34\) that $$ \left\\{w_{1}, v_{1}, \ldots, v_{n}\right\\} $$ is linearly dependent and also spans \(V\). By Lemma \(4.10\), one of the vectors in (1) is a linear combination of the preceding vectors. This vector cannot be \(w_{1}\), so it must be one of the \(v^{\prime}\) s, say \(v_{j}\). Thus by Problem \(4.34\), we can delete \(v_{j}\) from the spanning set (1) and obtain the spanning set $$ \left\\{w_{1}, v_{1}, \ldots, v_{j-1}, \quad v_{j+1} \ldots, v_{n}\right\\} $$ Now we repeat the argument with the vector \(w_{2}\). That is, because (2) spans \(V\), the set $$ \left\\{w_{1}, w_{2}, v_{1}, \ldots, v_{j-1}, \quad v_{j+1}, \ldots, v_{n}\right\\} $$ is linearly dependent and also spans \(V\). Again by Lemma \(4.10\), one of the vectors in (3) is a linear combination of the preceding vectors. We emphasize that this vector cannot be \(w_{1}\) or \(w_{2}\), because \(\left\\{w_{1}, \ldots, w_{m}\right\\}\) is independent; hence, it must be one of the \(v^{\prime}\) s, say \(v_{k} .\) Thus, by Problem \(4.34\), we can delete \(v_{k}\) from the spanning set (3) and obtain the spanning set $$ \left\\{w_{1}, w_{2}, v_{1}, \ldots, v_{j-1}, \quad v_{j+1}, \ldots, v_{k-1}, \quad v_{k+1}, \ldots, v_{n}\right\\} $$ We repeat the argument with \(w_{3}\), and so forth. At each step, we are able to add one of the \(w\) 's and delete one of the \(v\) 's in the spanning set. If \(m \leq n\), then we finally obtain a spanning set of the required form: $$ \left\\{w_{1}, \ldots, w_{m}, v_{i_{1}}, \ldots, v_{i_{n-\infty}}\right\\} $$ Finally, we show that \(m>n\) is not possible. Otherwise, after \(n\) of the above steps, we obtain the spanning set \(\left\\{w_{1}, \ldots, w_{n}\right\\} .\) This implies that \(w_{n+1}\) is a linear combination of \(w_{1}, \ldots, w_{n}\), which contradicts the hypothesis that \(\left\\{w_{i}\right\\}\) is linearly independent.

Let \(V\) be the external direct sum of vector spaces \(U\) and \(W\) over a field \(K\). (See Problem 4.76.) Let $$ \hat{U}=\\{(u, 0): u \in U\\} \quad \text { and } \quad \hat{W}=\\{(0, w): w \in W\\} $$ Show that (a) \(\hat{U}\) and \(\hat{W}\) are subspaces of \(V\), (b) \(V=\hat{U} \oplus \hat{W}\).

Show that the functions \(f(t)=\sin t, g(t) \cos t, h(t)=t\) from \(\mathbf{R}\) into \(\mathbf{R}\) are linearly independent. Set a linear combination of the functions equal to the zero function 0 using unknown scalars \(x, y, z\); that is, set \(x f+y g+z h=0\). Then show \(x=0, y=0, z=0\). We emphasize that \(x f+y g+z h=0\) means that, for every value of \(t\), we have \(x f(t)+y g(t)+z h(t)=0\). Thus, in the equation \(x \sin t+y \cos t+z t=0\) : \(\begin{aligned} \text { (i) Set } t=0 & \text { to obtain } & x(0)+y(1)+z(0)=0 & \text { or } & y=0 \\ \text { (ii) } \text { Set } t=\pi / 2 & \text { to obtain } & x(1)+y(0)+z \pi / 2=0 & \text { or } \quad x+\pi z / 2 &=0 . \\ \text { (iii) } & \text { Set } t=\pi & \text { to obtain } & x(0)+y(-1)+z(\pi)=0 & \text { or } \quad-y+\pi z &=0 \end{aligned}\) The three equations have only the zero solution; that is, \(x=0, y=0, z=0\). Thus, \(f, g, h\) are linearly independent.

Let \(V\) be the vector space of \(2 \times 2\) matrices over \(K\). Let \(W\) be the subspace of symmetric matrices. Show that \(\operatorname{dim} W=3\), by finding a basis of \(W\) Recall that a matrix \(A=\left[a_{i j}\right]\) is symmetric if \(A^{T}=A\), or, equivalently, each \(a_{i j}=a_{j i}\). Thus, \(A=\left[\begin{array}{ll}a & b \\\ b & d\end{array}\right]\) denotes an arbitrary \(2 \times 2\) symmetric matrix. Setting (i) \(a=1, b=0, d=0\); (ii) \(a=0, b=1, d=0\); (iii) \(a=0, b=0, \vec{d}=1\), we obtain the respective matrices: $$ E_{1}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right], \quad E_{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right], \quad E_{3}=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] $$ We claim that \(S=\left\\{E_{1}, E_{2}, E_{3}\right\\}\) is a basis of \(W\); that is, (a) \(S\) spans \(W\) and (b) \(S\) is linearly independent. (a) The above matrix \(A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]=a E_{1}+b E_{2}+d E_{3}\). Thus, \(S\) spans \(W\). (b) Suppose \(x E_{1}+y E_{2}+z E_{3}=0\), where \(x, y, z\) are unknown scalars. That is, suppose $$ x\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]+y\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]+z\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \quad \text { or } \quad\left[\begin{array}{ll} x & y \\ y & z \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] $$ Setting corresponding entries equal to each other yields \(x=0, y=0, z=0\). Thus, \(S\) is linearly independent. Therefore, \(S\) is a basis of \(W\), as claimed.
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