Problem 20 - Differential Equations and Linear Algebra

Question

Let \(V\) be an inner product space with basis \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right\\} .\) If \(\mathbf{x}\) and \(\mathbf{y}\) are vectors in \(V\) such that \(\left\langle\mathbf{x}, \mathbf{v}_{i}\right\rangle=\left\langle\mathbf{y}, \mathbf{v}_{i}\right\rangle\) for each \(i=1,2, \ldots, n,\) prove that \(\mathbf{x}=\mathbf{y}\).

Step-by-Step Solution

Answer

To prove that \(\mathbf{x} = \mathbf{y}\) under the given condition \(\langle\mathbf{x}, \mathbf{v}_i\rangle = \langle\mathbf{y}, \mathbf{v}_i\rangle\) for each \(i=1,2,\ldots,n\), we first expressed both vectors as linear combinations of the basis vectors. Then, we used the linearity of inner products and the orthogonality of basis vectors to show that the corresponding coefficients \(a_i\) and \(b_i\) of \(\mathbf{x}\) and \(\mathbf{y}\) are equal for each \(i\). Finally, we showed that since the coefficients are equal, \(\mathbf{x}\) and \(\mathbf{y}\) are indeed the same vector.

Step 1: Express \(\mathbf{x}\) and \(\mathbf{y}\) as linear combinations of basis vectors

Since \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\) is a basis for \(V\), we can express \(\mathbf{x}\) and \(\mathbf{y}\) as linear combinations of the basis vectors: \[\mathbf{x} = a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_n\mathbf{v}_n\] \[\mathbf{y} = b_1\mathbf{v}_1 + b_2\mathbf{v}_2 + \cdots + b_n\mathbf{v}_n\] where \(a_i\) and \(b_i\) are scalar coefficients.

Step 2: Use the linearity of inner products

Now we can use the linearity of inner products to express the condition \(\langle\mathbf{x}, \mathbf{v}_i\rangle = \langle\mathbf{y}, \mathbf{v}_i\rangle\) for each \(i=1,2,\ldots,n\): \[\langle a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_n\mathbf{v}_n, \mathbf{v}_i\rangle = \langle b_1\mathbf{v}_1 + b_2\mathbf{v}_2 + \cdots + b_n\mathbf{v}_n, \mathbf{v}_i\rangle\] Expanding this using the linearity of inner products gives: \[a_1\langle\mathbf{v}_1, \mathbf{v}_i\rangle + a_2\langle\mathbf{v}_2, \mathbf{v}_i\rangle + \cdots + a_n\langle\mathbf{v}_n, \mathbf{v}_i\rangle = b_1\langle\mathbf{v}_1, \mathbf{v}_i\rangle + b_2\langle\mathbf{v}_2, \mathbf{v}_i\rangle + \cdots + b_n\langle\mathbf{v}_n, \mathbf{v}_i\rangle\]

Step 3: Use orthogonality of basis vectors

Now we use the orthogonality of basis vectors to simplify the equation above. The inner product \(\langle\mathbf{v}_j, \mathbf{v}_i\rangle\) is zero if \(i \neq j\). Therefore, all terms in the sum except for those with \(i=j\) will vanish: \[a_i\langle\mathbf{v}_i, \mathbf{v}_i\rangle = b_i\langle\mathbf{v}_i, \mathbf{v}_i\rangle\] Since \(\mathbf{v}_i\) is non-zero and \(\langle\mathbf{v}_i, \mathbf{v}_i\rangle > 0\), we can divide both sides by \(\langle\mathbf{v}_i, \mathbf{v}_i\rangle\): \[a_i = b_i\]

Step 4: Prove \(\mathbf{x} = \mathbf{y}\)

Now that we have shown that each \(a_i\) is equal to the corresponding \(b_i\), we can prove that \(\mathbf{x} = \mathbf{y}\): \[\mathbf{x} = a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_n\mathbf{v}_n = b_1\mathbf{v}_1 + b_2\mathbf{v}_2 + \cdots + b_n\mathbf{v}_n = \mathbf{y}\] Hence, we have proven that \(\mathbf{x}=\mathbf{y}\) under the given condition.

Key Concepts

Linear Combinations

A linear combination is a way to express a vector using a set of basis vectors. Think of basis vectors as building blocks, like toy blocks you would use to build a tower. Similarly, any vector in the space can be built using these building blocks. In mathematical terms, if we have a set of basis vectors \( \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\} \), you can create any vector \( \mathbf{x} \) in that space by combining these basis vectors with some weights (or coefficients) like this: \[\mathbf{x} = a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_n\mathbf{v}_n\] Here, \(a_1, a_2, \ldots, a_n\) are the coefficients that scale the basis vectors. For another vector \( \mathbf{y} \), we can similarly express it as: \[\mathbf{y} = b_1\mathbf{v}_1 + b_2\mathbf{v}_2 + \cdots + b_n\mathbf{v}_n\] The beauty is that every vector in your space can be captured using these linear combinations, the coefficients just tell you how much of each piece—each basis vector—you need.

Orthogonality

In the context of inner product spaces, orthogonality tells us about the relationship between pairs of vectors. When two vectors are orthogonal, it means they are at right angles to each other—like the axes on a graph. But mathematically in an inner product space, this means their inner product is zero. For two vectors \( \mathbf{u} \) and \( \mathbf{v} \), this is expressed as: \[ \langle \mathbf{u}, \mathbf{v} \rangle = 0 \] Orthogonality is important when dealing with basis vectors. If your basis vectors are orthogonal, it simplifies many calculations. Specifically,

• The inner product of different basis vectors \( \langle \mathbf{v}_i, \mathbf{v}_j \rangle \) is zero if \( i eq j \).
• This means when you're working with sums of multiples of these vectors, all terms cancel out except when you're dealing with the same basis vector.

This makes checking whether two different expressed vectors are equal (like in our solution) much simpler because for a pair of orthogonal vectors, you're really only comparing the coefficients of the same vector. If each pair of coefficients are equal, the whole vectors are the same.

Basis Vectors

Basis vectors are a set of vectors in a vector space that, much like coordinates on a map, provide a reference to locate every point or vector in that space. In simpler terms, these are the minimal number of vectors that span a space. You can think of them as directions that you can follow using the right coefficients. Each basis vector points uniquely in a direction and is key to building all other vectors. Here's what makes them special:

• Spanning the Space: Any vector in the space can be expressed as a linear combination of basis vectors. This means they cover the entire space, no gaps.
• Linearly Independent: None of the basis vectors can be formed by a combination of others in the set. They're each bringing something unique to the table.

In our problem, we've considered vectors \( \mathbf{x} \) and \( \mathbf{y} \) using a set of basis vectors \( \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\} \). This means that knowing the relationship between these basis vectors allows us to explore and prove relations between any vectors built with them, such as \( \mathbf{x} \) and \( \mathbf{y} \). Ensuring that each combination has its unique coefficients offers a simplified way to equate and compare vectors within the space.