c7fe31a
committed
Commits
Comments (0)
Files changed (2)

+1 0changes.txt

+8 10src/sectionOD.xml
changes.txt
src/sectionOD.xml
<p>We begin with a proof by induction (<acroref type="technique" acro="I" />) of the first statement in the conclusion of the theorem. We use induction on the dimension of $V$ to show that if $\ltdefn{T}{V}{V}$ is a linear transformation, then there is a basis $B$ for $V$ such that the matrix representation of $T$ relative to $B$, $\matrixrep{T}{B}{B}$, is an upper triangular matrix.</p>
<p>To start suppose that $\dimension{V}=1$. Choose any nonzero vector $\vect{v}\in V$ and realize that $V=\spn{\set{\vect{v}}}$. Then we can determine $T$ uniquely by $\lt{T}{\vect{v}}=\beta\vect{v}$ for some $\beta\in\complexes$ (<acroref type="theorem" acro="LTDB" />). This description of $T$ also gives us a matrix representation relative to the basis $B=\set{\vect{v}}$ as the $1\times 1$ matrix with lone entry equal to $\beta$. And this matrix representation is upper triangular (<acroref type="definition" acro="UTM" />).</p>
+<p>To start suppose that $\dimension{V}=1$. Choose any nonzero vector $\vect{v}\in V$ and realize that $V=\spn{\set{\vect{v}}}$. Then $\lt{T}{\vect{v}}=\beta\vect{v}$ for some $\beta\in\complexes$, which determines $T$ uniquely (<acroref type="theorem" acro="LTDB" />). This description of $T$ also gives us a matrix representation relative to the basis $B=\set{\vect{v}}$ as the $1\times 1$ matrix with lone entry equal to $\beta$. And this matrix representation is upper triangular (<acroref type="definition" acro="UTM" />).</p>
<p>For the induction step let $\dimension{V}=m$, and assume the theorem is true for every linear transformation defined on a vector space of dimension less than $m$. By <acroref type="theorem" acro="EMHE" /> (suitably converted to the setting of a linear transformation), $T$ has at least one eigenvalue, and we denote this eigenvalue as $\lambda$. (We will remark later about how critical this step is.) We now consider properties of the linear transformation $\ltdefn{T\lambda I_V}{V}{V}$.</p>
<p>Define $W$ to be the subspace of $V$ that is the range of $T\lambda I_V$, $W=\rng{T\lambda I_V}$. We define a new linear transformation $S$, on $W$,
+<p>Let $W$ be the subspace of $V$ that is the range of $T\lambda I_V$, $W=\rng{T\lambda I_V}$, and define $k=\dimension{W}\leq m1$. We define a new linear transformation $S$, on $W$,
<p>Since $W$ is the range of $T\lambda I_V$, $\lt{\left(T\lambda I_V\right)}{\vect{w}}\in W$. And by <acroref type="property" acro="SC" />, $\lambda\vect{w}\in W$. Finally, applying <acroref type="property" acro="AC" /> we see by closure that the sum is in $W$ and so we conclude that $\lt{S}{\vect{w}}\in W$. This argument convinces us that it is legitimate to define $S$ as we did with $W$ as the codomain.</p>
<p>$S$ is a linear transformation defined on a vector space with dimension less than $m$, so we can apply the induction hypothesis and conclude that $W$ has a basis, $C=\set{\vectorlist{w}{k}}$, such that the matrix representation of $S$ relative to $C$ is an upper triangular matrix.</p>
+<p>$S$ is a linear transformation defined on a vector space with dimension $k$, less than $m$, so we can apply the induction hypothesis and conclude that $W$ has a basis, $C=\set{\vectorlist{w}{k}}$, such that the matrix representation of $S$ relative to $C$ is an upper triangular matrix.</p>
<p>By <acroref type="theorem" acro="DSFOS" /> there exists a second subspace of $V$, which we will call $U$, so that $V$ is a direct sum of $W$ and $U$, $V=W\ds U$. Choose a basis $D=\set{\vectorlist{u}{\ell}}$ for $U$. So $m=k+\ell$ by <acroref type="theorem" acro="DSD" />, and $B=C\cup D$ is basis for $V$ by <acroref type="theorem" acro="DSLI" /> and <acroref type="theorem" acro="G" />. $B$ is the basis we desire. What does a matrix representation of $T$ look like, relative to $B$?</p>
+<p>Beginning with the linearly independent set $C$, repeatedly apply <acroref type="theorem" acro="ELIS" /> to add vectors to $C$, maintaining a linearly independent set and spanning ever larger subspaces of $V$. This process will end with the addition of $mk$ vectors, which together with $C$ will span all of $V$. Denote these vectors as $D=\set{\vectorlist{u}{{mk}}}$. Then $B=C\cup D$ is a basis for $V$, and is the basis we desire for the conclusion of the theorem. So we now consider the matrix representation of $T$ relative to $B$.</p>
<p>Since the definition of $T$ and $S$ agree on $W$, the first $k$ columns of $\matrixrep{T}{B}{B}$ will have the upper triangular matrix representation of $S$ in the first $k$ rows. The remaining $\ell=mk$ rows of these first $k$ columns will be all zeros since the outputs of $T$ on $C$ are all contained in $W$. The situation for $T$ on $D$ is not quite as pretty, but it is close.</p>
+<p>Since the definition of $T$ and $S$ agree on $W$, the first $k$ columns of $\matrixrep{T}{B}{B}$ will have the upper triangular matrix representation of $S$ in the first $k$ rows. The remaining $mk$ rows of these first $k$ columns will be all zeros since the outputs of $T$ for basis vectors from $C$ are all contained in $W$ and hence are linear combinations of the basis vectors in $C$. The situation for $T$ on the basis vectors in $D$ is not quite as pretty, but it is close.</p>
<![CDATA[&=\vectrep{B}{\lt{T}{\vect{u}_i}+\zerovector}&&]]>\text{<acroref type="property" acro="Z" />}\\
<p>In the penultimate step of this proof, we have rewritten an element of the range of $T\lambda I_V$ as a linear combination of the basis vectors, $C$, for the range of $T\lambda I_V$, $W$, using the scalars $\scalarlist{a}{k}$. If we incorporate these $\ell$ column vectors into the matrix representation $\matrixrep{T}{B}{B}$ we find $\ell$ occurrences of $\lambda$ on the diagonal, and any nonzero entries lying only in the first $k$ rows. Together with the $k\times k$ upper triangular representation in the upper lefthand corner, the entire matrix representation is now clearly upper triangular. This completes the induction step, so for any linear transformation there is a basis that creates an upper triangular matrix representation.</p>
+<p>In the penultimate equality, we have rewritten an element of the range of $T\lambda I_V$ as a linear combination of the basis vectors, $C$, for the range of $T\lambda I_V$, $W$, using the scalars $\scalarlist{a}{k}$. If we incorporate these $mk$ column vectors into the matrix representation $\matrixrep{T}{B}{B}$ we find $mk$ occurrences of $\lambda$ on the diagonal, and any nonzero entries lying only in the first $k$ rows. Together with the $k\times k$ upper triangular representation in the upper lefthand corner, the entire matrix representation for $T$ is clearly upper triangular. This completes the induction step. So for any linear transformation there is a basis that creates an upper triangular matrix representation.</p>
<p>We have one more statement in the conclusion of the theorem to verify. The eigenvalues of $T$, and their multiplicities, can be computed with the techniques of <acroref type="chapter" acro="E" /> relative to any matrix representation (<acroref type="theorem" acro="EER" />). We take this approach with our upper triangular matrix representation $\matrixrep{T}{B}{B}$. Let $d_i$ be the diagonal entry of $\matrixrep{T}{B}{B}$ in row $i$ and column $i$. Then the characteristic polynomial, computed as a determinant (<acroref type="definition" acro="CP" />) with repeated expansions about the first column, is