---- > [!definition] Definition. ([[singular values]]) > Let [[inner product space]], $\big(W, \langle \star,\star \rangle\big)$ be finite-dimensional [[inner product space|inner product space]]s and suppose $T \in \text{Hom}(V,W)$. The **singular values** $\{ \sigma _{j} \}_{j=1}^{n}$ of $T$ are the [[eigenvalue|eigenvalues]] of $\sqrt{ T^{\dagger}T }$, with each [[eigenvalue]] $\lambda$ repeated [[dimension]] $E(\lambda,\sqrt{ T^{\dagger}T })$ times. [^1] > \ > **Remark.** Because they are [[eigenvalue|eigenvalues]] of a [[positive semidefinite operator]], **singular values** are always nonnegative; it is customary to list them in decreasing order:$\sigma_{1} \geq \sigma_{2} \geq \dots \geq \sigma _{m} \geq 0.$ > [!equivalence] Equivalences. >- [[singular values without taking the square root of an operator]] >- [[singular values are like eigenvalues of mismatched eigenvectors]] > [!basicproperties] > - > [[an operator has a number of singular values equal to the dimension of its source]] ; >- [[number of nonzero singular values equals rank]] ; >- $T$ has at most $\min(\dim V, \dim W)$ distinct singular values. [^1]: Recall that $E(\lambda,\sqrt{ T^{\dagger}T })$ denotes the $\lambda -$[[eigenspace]] of $\sqrt{T^{\dagger}T }$. > [!basicexample] > Define $T \in \endo({\rr^{4}})$ by $T(z_{1},z_{2},z_{3},z_{4})=(0,3z_{1},2z_{2},-3z_{4}).$ Let's find the [[singular values]] of $T$: \ **Step 1 Summary.** First we have to find $T^{\dagger}T$, and then $\sqrt{ T^{\dagger}T }$. We will take advantage of the fact that [[hom and mat are isomorphic]] and compute, all with respect to the standard [[basis]], $\MM(T)$ and $\MM(T^{\dagger})$, and then compute $\MM(T^{\dagger})\MM(T)$, which is by definition of [[matrix product|matrix multiplication]] equal to $\MM(T^{\dagger}T)$. Then we'll use [[linear maps act like matrix(-vector) multiplication]] to compute for arbitrary $v$ $\MM(T^{\dagger}T)\MM(v) = \MM(T^{\dagger}Tv)$ and deduce $T^{\dagger}Tv$ again via the [[hom and mat are isomorphic|isomorphism]]. \ **Step 1 Computation**. We have >- $Te_{1} = T(1,0,0,0)=(0,3,0,0)=0e_{1}+3e_{2}+0e_{3}+0e_{4}$; >- $Te_{2}=T(0,1,0,0)=(0,0,2,0)=0e_{1}+0e_{2}+2e_{3}+0e_{4}$; >- $Te_{3}=T(0,0,1,0)=(0,0,0,0)=0e_{1}+0e_{2}+0e_{3}+0e_{4}$; >- $Te_{4}=T(0,0,0,1)=(0,0,0,-3)=0e_{1}+0e_{2}+0e_{3}+-3e_{4}$. The first outputs above are tuples that pay no mind to any [[basis]]; the second outputs are [[vector]]s represented as [[linear combination]]s of the standard [[basis]]. \ We first compute $\begin{align} \MM(T)= & \begin{pmatrix} | & | & | & |\\ \MM(Te_{1}) & \MM(Te_{2}) & \MM(Te_{3}) & \MM(Te_{4}) \\ | & | & | & | \end{pmatrix} \\ = & \begin{pmatrix} 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix}. \end{align}$ Because [[adjoint matrix w.r.t. orthonormal bases equals conjugate transpose]], we have $\begin{align} \MM(T^{\dagger})= & \big(\MM(T)\big)^\top= & \begin{pmatrix} 0 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix}. \end{align}$ Now, $\MM(T^{\dagger}T)=\MM(T)\MM(T^{\dagger})=\begin{pmatrix} 9 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 9 \end{pmatrix}.$ Let's compute for arbitrary $v=z_{1}e_{1}+\dots+z_{4}e_{4}$ $\MM(T^{\dagger}Tv)$. We have $\MM(T^{\dagger}Tv)=\MM(T^{\dagger}T)\MM(v)=\begin{pmatrix} 9 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 9 \end{pmatrix} \begin{pmatrix} z_{1} \\ z_{2} \\ z_{3} \\ z_{4} \end{pmatrix},$ which yields $\MM(T^{\dagger}Tv)=\begin{pmatrix}9z_{1}\\4z_{2}\\0z_{3}\\9z_{4}\end{pmatrix}$. Now the definition of [[matrix]] tells us that these entries are the scalars required to write $\MM(T^{\dagger}Tv)$ in terms of the [[basis]]. Hence we have $T^{\dagger}Tv = 9z_{1}e_{1}+4z_{2}e_{2}+0z_{3}e_{3}+9z_{4}e_{4}$, which corresponds to the tuple $T^{\dagger}T(z_{1},z_{2},z_{3},z_{4})=(9z_{1},4z_{2},0,9z_{4})$. \ **Step 2 Summary**. Now we have to compute $\sqrt{ T^{\dagger}T }$ from $T^{\dagger}T$. $\MM(T^{\dagger}T)$ is [[upper-triangular matrix]], [[computing eigenvalues from upper-triangular matrix|hence its eigenvalues are]] $9$, $4$, and $0$. Moreover, it is a [[diagonal matrix]], so the standard [[basis]] is in fact an [[eigenbasis]]. This allows us to use the property discussed [[every positive semidefinite operator has exactly one positive square root|here]]. \ **Step 2 Computation.** From [[every positive semidefinite operator has exactly one positive square root|the property described here]] of $\sqrt{ T^{\dagger}T }$, since the standard [[basis]] happens to be an [[eigenbasis]] of $T^{\dagger}T$ we have $\begin{align} \sqrt{ T^{\dagger}T }(v)= & \sqrt{ T^{\dagger}T }(z_{1}e_{1} +\dots +z_{4}e_{4}) \\ = & z_{1}\sqrt{ T^{\dagger} T }e_{1} + \dots + z_{4} \sqrt{ T^{\dagger} T }e_{4} \\ = & z_{1}\sqrt{ 9 }e_{1} + z_{2}\sqrt{ 4 }e_{2} + z_{3} \sqrt{ 0 }e_{3} + z_{4} \sqrt{ 9 }e_{4}, \end{align}$ thus $\sqrt{ T^{\dagger} T }(z_{1},z_{2},z_{3},z_{4})=(3z_{1},2z_{2},0,3z_{4}).$ \ **Step 3 Summary.** We need to find the [[eigenvalue]]s of $\sqrt{ T^{\dagger}T }$ and the [[dimension|dimensions]] of their [[eigenspace|eigenspaces]]. \ **Step 3 Computation**. It is clear that the [[eigenvalue]]s are $3,2,0$, with corresponding [[eigenspace]] dimensions $2,1,1$. \ **ANSWER.** The [[singular values]] of $T$ are $3,3,2,0$. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```