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HomeMatlabWhat Is a Matrix Norm? – Nick Higham

What Is a Matrix Norm? – Nick Higham


A extra basic subordinate matrix norm might be outlined by taking totally different vector norms within the numerator and denominator:

notag   |A|_{alpha,beta} = displaystylemax_{xne 0}          frac{ |Ax|_{beta} }{ |x|_{alpha} }.

Some authors denote this norm by |A|_{alphatobeta}.

A helpful characterization of |A|_{alpha,beta} is given within the subsequent outcome. Recall that |cdot|^D denotes the twin of the vector norm |cdot|.

Theorem 1. For Ainmathbb{C}^{mtimes n},

notag    |A|_{alpha,beta} =           displaystylemax_{xinmathbb{C}^n atop y inmathbb{C}^m}           frac{mathrm{Re}, y^*Ax}x.

Proof. We’ve

notag begin{aligned}   displaystyle max_{xinmathbb{C}^n atop y inmathbb{C}^m}       frac{mathrm{Re}, y^*Ax}x  &=   max_{xinmathbb{C}^n} frac{1}x       max_{yinmathbb{C}^m} frac{mathrm{Re}, y^*Ax}_beta^D   &=   max_{xinmathbb{C}^n} frac Ax x  =  |A|_{alpha,beta}, end{aligned}

the place the second equality follows from the definition of twin vector norm and the truth that the twin of the twin norm is the unique norm.

We will now get hold of a connection between the norms of A and A^*. Right here, |A^*|_{beta^D,alpha^D} denotes max_{xne 0} |Ax|_alpha^D / |x|_beta^D.

Theorem 2. If Ainmathbb{C}^{mtimes n} then |A|_{alpha,beta} = |A^*|_{beta^D,alpha^D}.

Proof. Utilizing Theorem 1, we’ve

notag   |A^*|_{alpha,beta}    = displaystylemax_{xinmathbb{C}^n atop y inmathbb{C}^m}           frac{mathrm{Re}, y^*Ax}x    = displaystylemax_{xinmathbb{C}^n atop y inmathbb{C}^m}      frac{mathrm{Re}, x^*(A^*y)}{|x|_alpha |y|_{beta^D}}    = |A^*|_{beta^D,alpha^D}. quadsquare

If we take the alpha– and beta-norms to be the identical p-norm then we’ve |A|_p = |A^*|_q, the place p^{-1} + q^{-1} = 1 (giving, specifically, |A|_2 = |A^*|_2 and |A|_1 = |A^*|_infty, that are simply obtained instantly).

Now we give express formulation for the (alpha,beta) norm when alpha or beta is 1 or infty and the opposite is a basic norm.

Theorem 3. For Ainmathbb{C}^{mtimes n},

notag begin{alignedat}{2}       |A|_{1,beta} &= max_j | A(:,j) |_{beta},    &qquadqquad& (3)  |A|_{alpha,infty} &= max_i |A(i,:)^*|_{alpha}^D, &qquadqquad& (4)  end{alignedat}

For Ainmathbb{R}^{mtimes n},

notag    |A|_{infty,beta} = displaystylemax_{xin Z} |Az|_{beta},  qquadqquad (5)

the place

Z = { zinmathbb{R}^n: z_i = pm 1~mathrm{for~all}~i ,},

and if A is symmetric constructive semidefinite then

notag    |A|_{infty,1} = displaystylemax_{xin Z} z^T!Az.

Proof. For (3),

notag    |Ax|_{beta} = Big| sum_j x_j A(colon,j) Bigr|_{beta}         le displaystylemax_j | A(colon,j) |_{beta} sum_j |x_j|,

with equality for x=e_k, the place the utmost is attained for j=k. For (4), utilizing the Hölder inequality,

|Ax|_{infty} = displaystylemax_i | A(i,colon) x |      le max_i bigl( |A(i,colon)^*|_{alpha}^D |x|_{alpha} bigr)        =  |x|_{alpha} max_i |A(i,colon)^*|_{alpha}^D.

Equality is attained for an x that offers equality within the Hölder inequality involving the kth row of A, the place the utmost is attained for i=k.

Turning to (5), we’ve |A|_{infty,beta} = max__infty =1 |Ax|_beta. The unit dice {, xinmathbb{R}^n: -e le x le e,}, the place e = [1,1,dots,1]^T, is a convex polyhedron, so any level inside it’s a convex mixture of the vertices, that are the weather of Z. Therefore |x|_infty = 1 implies

notag   x = displaystylesum_{zin Z} lambda_z Z, quad mathrm{where} quad lambda_z ge 0,                                  quad sum_{zin Z} lambda_z = 1

after which

notag   |Ax|_beta = biggl| displaystylesum_{zin Z} lambda_z Az biggr|_beta   le max_{zin Z} |Az|_beta.

Therefore max__infty = 1 |Ax|_beta   le max_{zin Z} |Az|_beta, however trivially max_{zin Z} |Az|_beta le max__infty = 1 |Ax|_beta and (5) follows.

Lastly, if A is symmetric constructive semidefinite let xi_z = mathrm{sign}(Az) in Z. Then, utilizing a Cholesky factorization A = R^T!R (which exists even when A is singular) and the Cauchy–Schwarz inequality,

notag begin{aligned}   max_{zin Z} |Az|_1 &= max_{zin Z} xi_z^T Az                          = max_{zin Z} xi_z^T R^T!Rz                          &= max_{zin Z} (Rxi_z)^T Rz                          le max_{zin Z} |Rxi_z|_2 |Rz|_2                          &le max_{zin Z} |Rz|_2^2                          &= max_{zin Z} z^T!Az. end{aligned}

Conversely, for zin Z we’ve

notag    z^T!Az = |z^T!Az| le |z|^T|Az| = e^T |Az| = |Az|_1,

so max_{zin Z} z^T!Az le max_{zin Z} |Az|_1. Therefore max_{zin Z} z^T!Az = max_{zin Z} |Az|_1 = |A|_{infty,1}, utilizing (5). square

As particular instances of (3) and (4) we’ve

notag begin{aligned}   |A|_{1,infty} &= max_{i,j} |a_{ij}|, qquadqquadqquad (6)    |A|_{2,infty} &= max_i |A(i,:)^*|_2,    |A|_{1,2}      &= max_J |A(:,j)|_2. end{aligned}

We additionally get hold of by utilizing Theorem 2 and (5), for Ainmathbb{R}^{mtimes n},

notag   |A|_{2,1} = displaystylemax_{zin Z} |A^Tz|_2.

The (2,infty)-norm has lately discovered use in statistics (Cape, Tang, and Priebe, 2019), the motivation being that as a result of it satisfies

notag    |A|_{2,infty} le |A|_2 le m^{1/2} |A|_{2,infty},

the (2,infty)-norm might be a lot smaller than the 2-norm when m gg n and so generally is a higher norm to make use of in bounds. The (2,infty)– and (infty,2)-norms are utilized by Rebrova and Vershynin (2018) in bounding the 2-norm of a random matrix after zeroing a submatrix. They notice that the 2-norm of a random ntimes n matrix includes maximizing over infinitely many random variables, whereas the (infty,2)-norm and (2,infty)-norm contain solely 2^n and n random variables, respectively.

The (alpha,beta) norm shouldn’t be constant, however for any vector norm gamma, we’ve

notag begin{aligned}   |AB|_{alpha,beta}   &= max_{xne0} fracABxx   = max_{xne0} fracA(Bx)Bx                  fracBxx   le max_{xne0} fracA(Bx)Bx     max_{xne0} fracBxx   &le |A|_{gamma,beta} |B|_{alpha,gamma}. end{aligned}

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