Invertible Matrices And Change Of Basis

Hey guys! So, you're diving deep into linear algebra and pondering a question that's super common: is every invertible matrix a change of basis matrix? It's a fantastic question, and the short answer is, well, mostly yes, but with a crucial distinction tied to how we define these matrices in the first place. Let's break it down, because understanding this relationship is key to truly grasping the power of matrices in manipulating vector spaces. We're going to explore this topic in detail, ensuring you get a solid grip on the concepts, just like you'd expect from Plastik Magazine.

The Core Concept: What is a Change of Basis Matrix, Really?

First off, let's get on the same page about what a change of basis matrix actually is. In your course, you're likely treating it as the matrix of the identity operation from one basis to another. This is a super common and practical way to think about it. Imagine you have a vector space, and you're using two different sets of coordinates to describe vectors in that space – let's call them basis S and basis B. A change of basis matrix, often denoted as $P_{S o B}$ or similar, is the magic tool that allows you to convert the coordinates of a vector expressed in basis S into its equivalent coordinates in basis B. It essentially tells you how the old basis vectors (S) are represented in terms of the new basis vectors (B). If we have a vector $v$ with coordinates $[v]_S$ in basis S, and we want to find its coordinates $[v]_B$ in basis B, the relationship is given by: $[v]_B = P_{S o B} [v]_S$. This matrix $P_{S o B}$ is constructed by taking the coordinate vectors of the basis vectors of S (expressed in basis B) and stacking them up as columns. For example, if $S = \{s_1, s_2\}$ and $B = \{b_1, b_2\}$, and we express $s_1 = c_{11}b_1 + c_{21}b_2$ and $s_2 = c_{12}b_1 + c_{22}b_2$, then the change of basis matrix from S to B would be $P_{S o B} = \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix}$. The columns of this matrix are precisely the coordinates of the old basis vectors in terms of the new basis.

Now, the crucial part here is that for this to be a valid change of basis, the matrix must be invertible. Why? Because you need to be able to go back and forth between bases. If you can convert from S to B, you must be able to convert from B to S. This reversibility is guaranteed by the invertibility of the change of basis matrix. The inverse matrix, $P_{B o S}$, performs the conversion in the opposite direction. So, if a matrix represents a change of basis from S to B, it must be invertible. This leads us directly to the instructor's theorem and the heart of your question: if a matrix is invertible, does it always represent a change of basis? Let's dig into that.

The Instructor's Theorem and Its Implications

Your instructor introduced a theorem that likely states something along the lines of: The matrix of the identity operation from basis S to basis B is invertible, and its inverse is the matrix of the identity operation from basis B to basis S. This is a fundamental result in linear algebra. Let $I: V o V$ be the identity operation on a vector space $V$. If we choose basis S and basis B, the matrix of $I$ with respect to these bases is denoted as $[I]_{S}^{B}$ (or often just $P_{S o B}$), representing the transformation of coordinates from basis S to basis B. The theorem essentially confirms that if you can express the identity operation as a transformation between two bases, the matrix representing this transformation is indeed invertible. This invertibility is a direct consequence of the fact that you are dealing with legitimate bases for the same vector space. If S and B are bases for the same $n$-dimensional vector space, then any change of basis matrix between them will be an $n imes n$ matrix. The existence of a basis transformation implies that the transformation itself must be bijective (one-to-one and onto), which in turn means its matrix representation must be invertible.

Consider the identity map. When we apply it to a vector $v$, we get $v$ back. So, $[I(v)]_B = [v]_B$ and $[I(v)]_S = [v]_S$. The matrix of the identity map from S to B, $[I]_{S}^{B}$, when applied to the coordinate vector $[v]_S$, should produce the coordinate vector $[v]_B$. That is, $[v]_B = [I]_{S}^{B} [v]_S$. Similarly, the matrix of the identity map from B to S, $[I]_{B}^{S}$, should satisfy $[v]_S = [I]_{B}^{S} [v]_B$. If we substitute the second equation into the first, we get $[v]_B = [I]_{S}^{B} ([I]_{B}^{S} [v]_B) = ([I]_{S}^{B} [I]_{B}^{S}) [v]_B$. Since this must hold for all vectors $v$ (and thus for all coordinate vectors $[v]_B$), it must be that $[I]_{S}^{B} [I]_{B}^{S} = I$, the identity matrix. This proves that $[I]_{S}^{B}$ is invertible and its inverse is $[I]_{B}^{S}$. The theorem is solid. It guarantees that any matrix that is a change of basis matrix (i.e., represents the identity operation between two bases) is invertible.

The Converse: Is Every Invertible Matrix a Change of Basis?

Now, this is where the nuance comes in, and it's the crux of your question. The theorem states that if a matrix is a change of basis matrix, then it's invertible. The question is, if we have an invertible matrix, does it automatically mean it is a change of basis matrix? The answer is yes, provided we can establish the context of two bases for the same vector space.

Let's say you have an $n imes n$ invertible matrix $A$. We are working in an $n$-dimensional vector space $V$. Let $B = \{b_1, b_2, ..., b_n\}$ be any chosen basis for $V$. We can use the invertible matrix $A$ to define a new set of vectors, let's call them $c_1, c_2, ..., c_n$, such that the coordinate vectors of these new vectors in basis B are the columns of A. Specifically, let $[c_i]_B$ be the $i$-th column of $A$. Since $A$ is invertible, its columns are linearly independent. Therefore, the set of vectors $C = \{c_1, c_2, ..., c_n\}$ forms a linearly independent set of $n$ vectors in an $n$-dimensional space. This means $C$ is also a basis for $V$. Now, the matrix $A$ can be interpreted as the change of basis matrix from basis $C$ to basis $B$. That is, $A = P_{C o B}$. Why? Because the definition of $P_{C o B}$ is the matrix whose columns are the coordinate vectors of the old basis vectors (C) expressed in terms of the new basis (B). And that's exactly how we constructed it: the columns of $A$ are the coordinate vectors of $c_i$ in basis B.

So, given any invertible $n imes n$ matrix $A$ and any basis $B$ for an $n$-dimensional vector space $V$, we can construct a new basis $C$ such that $A$ becomes the change of basis matrix from $C$ to $B$. Alternatively, we could fix one basis, say $S$, and use the invertible matrix $A$ to define a linear transformation $T$ such that $[T]_S = A$. If $A$ is invertible, then $T$ is an isomorphism. We can then consider the transformation of basis. If we have a standard basis $E$, and an invertible matrix $A$, we can define a new basis $B$ such that $A$ is the change of basis matrix from the standard basis $E$ to the new basis $B$. The columns of $A$ would be the vectors of the new basis $B$ expressed in the standard basis $E$. So, $[v]_B = A [v]_E$ is not quite right. It's $[v]_E = A [v]_B$. This means $A$ is the change of basis matrix from $B$ to $E$, i.e., $A = P_{B o E}$. The columns of $A$ are the vectors of the basis $B$ (expressed in the standard basis $E$). Yes, that's it!

The Crucial Caveat: Context is Everything!

The key takeaway here, guys, is that the interpretation of an invertible matrix as a change of basis matrix depends on the context. An invertible matrix $A$ doesn't inherently contain information about two specific bases. Rather, it enables a change of basis between some pair of bases. If you are given an invertible matrix $A$, you can always find two bases, say $B$ and $C$, for the same vector space such that $A$ is the change of basis matrix from $C$ to $B$ (or $B$ to $C$, depending on convention). However, an arbitrary invertible matrix $A$ doesn't automatically come paired with two pre-existing bases for which it serves as the bridge. The mathematician's perspective is that an invertible $n imes n$ matrix represents an isomorphism from an $n$-dimensional vector space to itself. This isomorphism can always be realized as a change of basis between some pair of bases.

Let's re-iterate the instructor's definition: