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A standard real Gaussian random matrix has independent and identically distributed standard normal entries. Each column is an orthogonally invariant random vector: its probability density function is proportional to $\exp (-\Vert z{\Vert }_2^2/2)$. Consequently multiplying on either side by a fixed orthogonal matrix preserves the joint distribution. Its singular values are random; the expressions $\sqrt{m}\pm \sqrt{n}$ for an $m\times n$ matrix describe useful spectral scales, not deterministic endpoints that hold for every realization.

For an orthogonal matrix $A$, invariance in distribution givesThe orthogonal group is transitive on each Euclidean sphere, so ${\phi }_X(u)$ depends only on $\Vert u{\Vert }^2$. DefineThe continuity of a characteristic function makes $f$ continuous. Taking $A=-I$ also gives $X\stackrel{d}{=}-X$, and henceso $f$ is real-valued. Thusas for every orthogonally invariant random vector.

Write $\ell =r/2$; the requested expression requires the positive even integer $r\ge 2$. Split the thin singular value decomposition after $\ell$ columns:Here $\Vert D_b{\Vert }_2={\sigma }_{\ell +1}=\Vert A-A_{\ell }{\Vert }_2$. Each column of a Gaussian random matrix is an orthogonally invariant random vector: its probability density function is proportional to $\exp (-\Vert z{\Vert }_2^2/2)$, and an orthogonal matrix preserves both $\Vert z{\Vert }_2$ and volume. Thus $V^{T}X$, where $V=[V_a,V_b]$, again has independent and identically distributed standard normal entries. In particular both displayed blocks have the Gaussian distributions needed below.

First prove a deterministic error bound for a randomized range finder. Put $P=Q{Q}^T$ and suppose ${\Omega }_a$ has full row rank. Its Moore-Penrose pseudoinverse is ${\Omega }_a^{†}={\Omega }_a^T({\Omega }_a{\Omega }_a^T{)}^{-1}$, so ${\Omega }_a{\Omega }_a^{†}=I_{\ell }$. The QR decomposition implies $(I_m-P)AX=0$, henceMultiplication by ${\Omega }_a^{†}$ expresses the missed leading columns in terms of the trailing columns. Substitution into $(I_m-P)A$ gives the exact identityFor any matrix $F$, the spectral norm of $[-FI]$ is $\sqrt{1+\Vert F{\Vert }_2^2}$, because its product with its transpose is $F{F}^T+I$. Using submultiplicativity of the operator norm, the orthonormality of $U_b$, and the contraction property of the orthogonal projection, we obtainThis estimate is deterministic and also covers rank-deficient $A$; if ${\sigma }_{\ell +1}=0$, its right-hand side vanishes.

Now use the paper's permitted singular value estimates. Apply the lower estimate to the tall matrix ${\Omega }_a^T$ of size $r\times \ell$, rather than to its wide transpose. It givesThe upper estimate for ${\Omega }_b$, transposing if necessary, givesThus, with $K=2/(1-1/\sqrt{2})$,Since $n/r>1$, the deterministic estimate yieldsThis proves the requested conclusion under the explicitly authorized estimates.

The probabilistic qualification is necessary. A Gaussian random matrix does not always have all its singular values between the two endpoints printed in the hint. Already a $1\times 1$ example would require $|Z|\le 2$ for every standard normal random variable $Z$, which fails with positive probability. The proof above holds on the event where the two block estimates hold. In a probabilistic formulation one supplies estimates for that event, usually allowing a positive margin in both endpoints; no failure probability is specified by the paper.

Nor can the requested inequality with one universal $C$ hold almost surely for every $A$. For example, take $n=m=3$, $r=2$, $A=\mathrm{diag}(L,1,1)$ with $L>1$, and a sampling matrixAt $X_0$, the sampled column space misses the first coordinate, so $\Vert A-\hat{A}{\Vert }_2=L$, whereas $\Vert A-A_1{\Vert }_2=1$. Choose $L>C\sqrt{3/2}$. Because $A{X}_0$ has full column rank, its orthogonal projection depends continuously on $X$ near $X_0$ through $AX(X^T{A}^{T}AX{)}^{-1}{X}^T{A}^T$. The strict violation persists in an open neighborhood of $X_0$. The joint Gaussian probability density function is positive there, so that neighborhood has positive probability. This identifies the missing qualification without weakening the conditional calculation.

An orthogonally invariant random vector has a probability measure determined by its radial pushforward measure. Here $\sigma$ is normalized surface measure on a sphere, and the formula holds for bounded measurable $f$ and for nonnegative measurable $f$. It includes an atom of a measure at zero: the inner integral there is simply $f(0)$.

To prove the formula first for continuous bounded $f$, average $f(Qx)$ over normalized Haar measure of the compact orthogonal group. Right invariance makes this average depend only on $|x|$; averaging also over surface measure on a sphere identifies it with the spherical average. Invariance of $\mu$ and Fubini's theorem then give the identity. Uniqueness of finite Borel measures and measurable approximation extend it to the stated classes of $f$. No extra sphere-area factor belongs in an average against normalized surface measure on a sphere.